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Quadratic Equations in One Variables.pptx

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The document discusses solving quadratic equations in one variable of the form ax^2 + bx + c = 0. It provides examples of quadratic equations and shows how to rewrite them in standard form. It then covers methods for solving quadratic equations, including extracting square roots, factoring, completing the square, and using the quadratic formula. It also discusses the nature of the roots based on the discriminant and provides rules for determining the sum and product of the roots.

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Quadratic Equation in One Variable
Examples: ï±đ‘„2 + 5đ‘„ + 3 = 0 2đ‘„2 − 4đ‘„ + 9 = 0 ï±đ‘„ 3 + đ‘„2 2 = 5  đ‘„ − 2 đ‘„ + 3 = 0 ï±đ‘„2 − 5 = đ‘„ + 6 ï±đ‘„ đ‘„ − 4 = 1 Quadratic Equation in One Variable A quadratic equation in one variable is of the form đ‘Žđ‘„2 + đ‘đ‘„ + 𝑐 = 0, where 𝑎, 𝑏, and 𝑐 are real number coefficients and 𝑎 ≠ 0. This equation is in the second degree wherein the highest exponent is 2.
đ‘„ − 2 đ‘„ + 3 = 0 Rewrite each equation in standard form. Then, identify the value of 𝒂, 𝒃, and 𝒄. 1. đ‘„ − 2 đ‘„ + 3 = 0 F O I L F: đ‘„2 O: 3đ‘„ I: −2đ‘„ L: −6 Quadratic Equation: đ‘„2 − đ‘„ − 6 = 0 Therefore, 𝑎 = 1; 𝑏 = −1; 𝑐 = −6.
Rewrite each equation in standard form. Then, identify the value of 𝒂, 𝒃, and 𝒄. 2. đ‘„2 − 5 = đ‘„ + 6 Quadratic Equation: đ‘„2 − đ‘„ + 1 = 0 đ‘„2 − 5 = đ‘„ + 6 đ‘„2 − 5 − đ‘„ = đ‘„ − 6 đ‘„2 − đ‘„ − 5 + 6 = −6 + 6 đ‘„2 − đ‘„ + 1 Therefore, 𝑎 = 1; 𝑏 = −1; 𝑐 = 1.
Rewrite each equation in standard form. Then, identify the value of 𝒂, 𝒃, and 𝒄. 3. (2đ‘„ + 1)2 = 5đ‘„ Quadratic Equation: 4đ‘„2 − đ‘„ + 1 = 0 (2đ‘„ + 1)2 = 5đ‘„ 4đ‘„2 + 4đ‘„ + 1 = 5đ‘„ 4đ‘„2 + 4đ‘„ − 5đ‘„ + 1 = 5đ‘„ − 5đ‘„ 4đ‘„2 − đ‘„ + 1 = 0 Therefore, 𝑎 = 4; 𝑏 = −1; 𝑐 = 1.
Rewrite each equation in standard form. Then, identify the value of 𝒂, 𝒃, and 𝒄. DRILL 1. (đ‘„ + 5)2 = 3 2. đ‘„ đ‘„ + 2 = 5đ‘„2 − 4 3. 2đ‘„ + 1 đ‘„ − 3 = đ‘„ + 1 4. đ‘„ − 4 đ‘„ − 6 = 2 5. 3 4 đ‘„2 − 2 3 đ‘„ = 3 5
The Roots of Quadratic Equations
Nature of the Roots If a, b, and c are real numbers and đ· = 𝑏2 − 4𝑎𝑐, then the roots of the quadratic equation đ‘Žđ‘„2 + đ‘đ‘„ + 𝑐 = 0 are: I. real and unequal, if D > 0; II. real and equal, if D = 0; III. imaginary and unequal, if D < 0; Moreover, the roots are: IV. rational and unequal, if D is a perfect square; V. irrational and unequal, if D is not a perfect square. Discriminant (𝒃𝟐 − 𝟒𝒂𝒄) The discriminant provided information regarding the nature of the roots of the quadratic equation.
Determine the nature of the roots of each equation. 1. đ‘„2 − 6đ‘„ + 9 = 0 đ‘„2 − 6đ‘„ + 9 = 0; a = 1, b = -6, and c = 9 đ· = 𝑏2 − 4𝑎𝑐 đ· = (−6)2 −4(1)(9) đ· = 36 − 36 đ· = 0 Therefore, the roots of 𝒙𝟐 − 𝟔𝒙 + 𝟗 = 𝟎 are real, equal, and rational.
Determine the nature of the roots of each equation. 2. 2đ‘„2 − 4đ‘„ + 5 = 0 2đ‘„2 − 4đ‘„ + 5 = 0; a = 2, b = -4, and c = 5 đ· = 𝑏2 − 4𝑎𝑐 đ· = (−4)2 −4(2)(5) đ· = 16 − 40 đ· = −24 Therefore, the roots of 𝟐𝒙𝟐 − 𝟒𝒙 + 𝟓 = 𝟎 are imaginary and unequal.
Determine the nature of the roots of each equation. DRILL 1. đ‘„2 + 10đ‘„ + 22 = 0 2. 4đ‘„2 + 2đ‘„ + 4 = 0 3. 2đ‘„2 − 4đ‘„ − 4 = 0 4. đ‘„2 − 10đ‘„ + 22 = 0 5. 3 4 đ‘„2 − 2 3 đ‘„ − 3 5 = 0
Sum and Product of the Roots Sum of The Roots = − 𝑏 𝑎 ; Product of the Roots = 𝑐 𝑎 Find the sum and product of the roots of the following equations. 1. đ‘„2 − 6đ‘„ + 9 = 0 đ‘„2 − 6đ‘„ + 9 = 0; a = 1, b = -6, and c = 9 𝑆𝑈𝑀 = −𝑏 𝑎 đ‘ƒđ‘…đ‘‚đ·đ‘ˆđ¶đ‘‡ = 𝑐 𝑎 = −(−6) 1 = 6 = 6 1 = 9 1 = 9
Find the sum and product of the roots of the following equations. 2. 2đ‘„2 − 4đ‘„ + 5 = 0 2đ‘„2 − 4đ‘„ + 5 = 0; a = 2, b = -4, and c = 5 𝑆𝑈𝑀 = −𝑏 𝑎 đ‘ƒđ‘…đ‘‚đ·đ‘ˆđ¶đ‘‡ = 𝑐 𝑎 = 2 = 4 2 = −(−4) 2 = 2.5 = 5 2
Determine the sum and products of the roots of each equation. DRILL 1. đ‘„2 + 10đ‘„ + 22 = 0 2. 4đ‘„2 + 2đ‘„ + 4 = 0 3. 2đ‘„2 − 4đ‘„ − 4 = 0 4. đ‘„2 − 10đ‘„ + 22 = 0 5. 3 4 đ‘„2 − 2 3 đ‘„ − 3 5 = 0
Solving Quadratic Equations by: Extracting Square Roots Factoring Completing The Square Using The Quadratic Formula
No. of Solutions in a Quadratic E quation  Every quadratic equation in one variable has exactly two solutions or roots.
Square Root Property  If đ‘„2 = 𝑛, then đ‘„ = 𝑛 or đ‘„ = − 𝑛.
‱ If 𝑏 = 0, the equation is a pure or incomplete quadratic equation. ‱ Example: đ‘„2 − 4 = 0 and 3đ‘„2 + 6 = 0 ‱ If 𝑏 ≠ 0, the equation is a complete quadratic equation. ‱ Example: đ‘„2 − 4đ‘„ + 1 = 0 and 3đ‘„2 + 2đ‘„ − 1 = 0 A. Extracting Square Roots Step 1: Express the quadratic equation in standard form. Step 2: Factor the quadratic expression. Step 3: Apply the zero-product property and set each variable factor equal to 0. Step 4: Solve the resulting linear equations.
Solve the quadratic equation by extracting the square root. 1. đ‘„2 − 11 = 0 đ‘„2 − 11 = 0 đ‘„2 = 11 đ‘„ = ± 11 To Check: If đ‘„ = 11, If đ‘„ = − 11 ( 11)2 −11 = 0 − 11 2 − 11 = 0 11 − 11 = 0 11 − 11 = 0 0 = 0 True 0 = 0 True
Solve the quadratic equation by extracting the square root. 2. 2đ‘„2 + 3 = 39 2đ‘„2 + 3 = 39 2đ‘„2 = 36 đ‘„2 = 18 đ‘„ = ± 18 đ‘„ = ± (9)(2) đ‘„ = ± 9 2 đ‘„ = ±3 2 To Check: If đ‘„ = 2 3, If đ‘„ = −3 2 2(3 2)2 + 3 = 39 −3 2 2 + 3 = 39 2(9)(2) + 3 = 39 2 9 2 + 3 = 39 36 + 3 = 39 36 + 3 = 39 39 = 39 True 39 = 39 True
Solve the quadratic equation by extracting the square root. 3. (đ‘„ + 4)2 = 1 (đ‘„ + 4)2= 1 đ‘„ + 4 2 = 1 đ‘„ + 4 = ±1 To Check: If đ‘„ = −3, If đ‘„ = −5 (−3 + 4)2 = 1 −5 + 4 2 = 1 (1)2= 1 (−1)2= 1 1 = 1 True 1 = 1 True For đ‘„ + 4 = 1, đ‘„ = −3 For đ‘„ + 4 = −1, đ‘„ = −5
Solve the following quadratic equations using the Square Root Property. DRILL 1. 6đ‘„2 − 24 = 0 2. 2đ‘„2 − 34 = 0 3. (đ‘„ + 2)2 = 9 4. (4đ‘„ + 1)2 = 196
Zero Property of Equality  The product đŽđ” = 0, if and only if 𝐮 = 0 or đ” = 0.
B. Factoring Step 1: Clear all fractions (if any) and write the given equation in the form đ‘Žđ‘„2 + đ‘đ‘„ + 𝑐 = 0. Step 2: Factor the product of the coefficient of đ‘„ and the constant term of the given quadratic equation. Step 3: Express the coefficient of the middle term as the sum or difference of the factors obtained in Step 2. Step 4: Use zero product property to put each linear factor equal to 0. Step 5: Solve the resulting linear equations.
Solve the quadratic equation by factoring. 1. đ‘„2 − 9đ‘„ + 18 = 0 đ‘„2 − 9đ‘„ + 18 = 0 đ‘„ − 3 đ‘„ − 6 = 0 đ‘„ − 3 = 0 đ‘„ − 6 = 0 đ‘„ = 3 đ‘„ = 6 TO CHECK: đ‘„ = 3 đ‘„2 − 9đ‘„ + 18 = 0 32 − 9 3 + 18 = 0 9 − 27 + 18 = 0 9 − 9 = 0 0 = 0 True TO CHECK: đ‘„ = 6 đ‘„2 − 9đ‘„ + 18 = 0 62 − 9 6 + 18 = 0 36 − 54 + 18 = 0 36 − 36 0 = 0 True
Solve the quadratic equation by factoring. 2. đ‘„2 + 12 = 7đ‘„ đ‘„2 + 12 = 7đ‘„ đ‘„2 − 7đ‘„ + 12 = 0 đ‘„ − 3 đ‘„ − 4 = 0 đ‘„ − 3 = 0 đ‘„ − 4 = 0 đ‘„ = 3 đ‘„ = 4 TO CHECK: đ‘„ = 3 đ‘„2 + 12 = 7đ‘„ 32 + 12 = 7(3) 9 + 12 = 21 21 = 21 True TO CHECK: đ‘„ = 4 đ‘„2 + 12 = 7đ‘„ 42 + 12 = 7(4) 16 + 12 = 28 28 = 28 True
Solve the following quadratic equations by factoring. DRILL 1. đ‘„2 − 5đ‘„ + 6 = 0 2. đ‘„2 − 3đ‘„ − 18 = 0 3. 2đ‘„2 + 3đ‘„ − 2 = 0 4. 2đ‘„2 + 7đ‘„ − 4 = 0
C. Completing the Square Step 1: Rewrite the equation in the form đ‘Žđ‘„2 + đ‘đ‘„ + 𝑐 = 0. Step 2: Add to both sides the term needed to complete the square. Step 3: Factor the perfect square trinomial. Step 4: Solve the resulting equation by using the square root property.. ‱ To make an expression of the form đ‘„2 + đ‘đ‘„ a perfect square trinomial, add 𝑏 2 2 to it.
Solve the quadratic equation by completing the square. 1. đ‘„2 + 6đ‘„ + 8 = 0 đ‘„2 + 6đ‘„ + 8 = 0 đ‘„2 + 6đ‘„ = −8 đ‘„2 + 6đ‘„ + 6 2 2 = −8 đ‘„2 + 6đ‘„ + (3)2= −8 đ‘„2 + 6đ‘„ + 9 = −8 + 9 đ‘„ + 3 2 = 1 (đ‘„ + 3)2= ± 1
Solve the quadratic equation by completing the square. 1. đ‘„2 + 6đ‘„ + 8 = 0 đ‘„ + 3 = ±1 đ‘„ = −3 ± 1 đ‘„ = −3 + 1 = −2 or đ‘„ = −3 − 1 = −4 TO CHECK: đ‘„ = −2 đ‘„2 + 6đ‘„ + 8 = 0 −22 + 6 −2 + 8 = 0 4 − 12 + 8 = 0 4 − 4 = 0 0 = 0 True TO CHECK: đ‘„ = −4 đ‘„2 + 6đ‘„ + 8 = 0 −42 + 6 −4 + 8 = 0 16 − 24 + 8 = 0 16 − 16 = 0 0 = 0 True SOLUTIONS: −2 & − 4
Solve the quadratic equation by completing the square. 2. đ‘„2 − 2đ‘„ = 1 đ‘„2 − 2đ‘„ = 1 đ‘„2 − 2đ‘„ + 2 2 2 = 1 đ‘„2 − 2đ‘„ + 1 2 = 1 đ‘„2 − 2đ‘„ + 1 = 1 + 1 (đ‘„ − 1)2= 2 (đ‘„ − 1)2= ± 2 đ‘„ = 1 + 2 or đ‘„ = 1 − 2 SOLUTIONS:1 + 2 & 1 − 2
Solve the quadratic equation by completing the square. 3. 4đ‘„2 − 5đ‘„ − 6 = 0 4đ‘„2 − 5đ‘„ − 6 = 0 4đ‘„2 − 5đ‘„ = 6 4 (đ‘„2 − 5 4 đ‘„) = 6 4 (đ‘„2 − 5 4 đ‘„) 4 = 6 4 đ‘„2 − 5 4 đ‘„ = 3 2
Solve the quadratic equation by completing the square. 3. 4đ‘„2 − 5đ‘„ − 6 = 0 Complete the Square: 5 4 2 2 = 5 8 2 = 25 64 đ‘„2 − 5 4 đ‘„ + 25 64 = 3 2 + 25 64 Add the fractions on the right side. 3 2 + 25 64 = 96 64 + 25 64 = 121 64
Solve the quadratic equation by completing the square. 3. 4đ‘„2 − 5đ‘„ − 6 = 0 đ‘„2 − 5 4 đ‘„ + 25 64 = 121 64 đ‘„ − 5 8 2 = 121 64 đ‘„ − 5 8 2 = ± 121 64
Solve the quadratic equation by completing the square. 3. 4đ‘„2 − 5đ‘„ − 6 = 0 đ‘„ − 5 8 = ± 11 8 đ‘„ = 5 8 ± 11 8 đ‘„ = 5 8 + 11 8 = 16 8 = 2 đ‘„ = 5 8 − 11 8 = − 6 8 = − 3 4 SOLUTIONS: 2 & − 3 4
Solve the following quadratic equations by completing the square. DRILL 1. đ‘„2 − 2đ‘„ − 8 = 0 2. đ‘„2 + 7đ‘„ − 18 = 0 3. 3đ‘„2 = 2 − đ‘„ 4. đ‘„2 = 9
D. Using the Quadratic Formula đ‘„ = −𝑏 ± 𝑏2 − 4𝑎𝑐 2𝑎 đ‘„ = −𝑏 ± 𝑏2 − 4𝑎𝑐 2𝑎 đ‘„ = −9 ± 92 − 4(1)(14) 2(1) Solve the quadratic equation using the quadratic formula. 1. đ‘„2 + 9đ‘„ + 14 = 0
Solve the quadratic equation using the quadratic formula. 1. đ‘„2 + 9đ‘„ + 14 = 0 đ‘„ = −9 ± 81 − 56 2 đ‘„ = −9 ± 25 2 đ‘„ = −9 ± 5 2
Solve the quadratic equation using the quadratic formula. 1. đ‘„2 + 9đ‘„ + 14 = 0 đ‘„ = −9 ± 5 2 đ‘„ = −9+5 2 or đ‘„ = −9−5 2 đ‘„ = −4 2 or đ‘„ = −14 2 đ‘„ = −2 or đ‘„ = −7
Solve the following quadratic equations using the quadratic formula. DRILL 1. đ‘„2 − 2đ‘„ + 1 = 0 2. đ‘„2 − 6đ‘„ + 9 = 0 3. đ‘„2 − 5đ‘„ − 5 = 0 4. 3đ‘„2 − 5đ‘„ − 2 = 0
Equations in Quadratic Form (𝑎𝑓2 + 𝑏𝑓 + 𝑐 = 0)
Equations in Quadratic Form An equation in quadratic form is any equation of the form 𝑎𝑓2 + 𝑏𝑓 + 𝑐 = 0, where 𝑓 is any algebraic expression. These equations can be solved using simple substitution. Example: Solve the following equations. 1. 9đ‘„4 − 52đ‘„2 + 64 = 0 In 9(đ‘„2 )2 −52 đ‘„2 + 64 = 0, let 𝑓 = đ‘„2 . 9𝑓2 − 52𝑓 + 64 = 0 𝑓 − 4 9𝑓 − 16 = 0 𝑓 − 4 = 0 or 9𝑓 − 16 = 0 𝑓 = 4 or 𝑓 = 16 9
Example: Solve the following equations. 1. 9đ‘„4 − 52đ‘„2 + 64 = 0 Let us solve for đ‘„. đ‘„2 = 𝑓 = 4 or đ‘„2 = 𝑓 = 16 9 đ‘„2 = 4 or đ‘„2 = 16 9 đ‘„ = ± 4 or đ‘„ = ± 16 9 đ‘„ = ±2 or đ‘„ = ± 4 3 Solution Set: −2, − 4 3 , 4 3 , 2
Example: Solve the following equations. 2. (2đ‘„ − 1)2 − 6 2đ‘„ − 1 = −8 In(2đ‘„ − 1)2−6 2đ‘„ − 1 = −8, let 𝑔 = 2đ‘„ − 1. 𝑔2 − 6𝑔 = −8 𝑔2 − 6𝑔 + 8 = 0 𝑔 − 4 𝑔 − 2 = 0 𝑔 − 4 = 0 or 𝑔 − 2 = 0 𝑔 = 4 or 𝑔 = 2
Example: Solve the following equations. 2. (2đ‘„ − 1)2 − 6 2đ‘„ − 1 = −8 Let us solve for đ‘„. 2đ‘„ − 1 = 𝑔 = 4 or 2đ‘„ − 1 = 𝑔 = 2 2đ‘„ − 1 = 4 or 2đ‘„ − 1 = 2 2đ‘„ = 5 or 2đ‘„ = 3 đ‘„ = 5 2 or đ‘„ = 3 2 Solution Set: 5 2 , 3 2
Solve for 𝒙 in each equation. DRILL 1. đ‘„4 + 4 = 5đ‘„2 2. đ‘„4 − 17đ‘„2 + 16 = 0
Rational Algebraic Equations Transformable to Quadratic Equation
Rational Algebraic Equations Rational algebraic equations are equations that contain one or more rational algebraic expressions (algebraic expressions in the form 𝑛 𝑑 , where 𝑛 and 𝑑 are polynomials and 𝑑 ≠ 0). These equations can be reduced to polynomial equations by clearing off fractions. Examples: 5đ‘„ 6 − 1 2 = 1 3 đ‘„ 2 + 3đ‘„ = 7 4đ‘„âˆ’6 2đ‘„âˆ’3 = 7 đ‘„+1
But, take note, not all rational algebraic equations can be transformed into quadratic equations. Rational Algebraic Equations Transformable to Quadratic Equations 1. It has to be a rational equation. 2. There should only be one variable. 3. If the expressions are linear, then there should be a variable on both numerator and denominator. Example: 4đ‘„âˆ’6 2đ‘„âˆ’3 4. If the denominators are all constant, then the numerator should have at least one variable that has a power of 2. Example: đ‘„2+1 7
Solving Rational Algebraic Equations Transformable to Quadratic Equations by Finding the LCD 1. Multiply both sides of the equation by the LCD. 2. Apply the distributive law. 3. Combine like terms. 4. Apply the inverse operation. 5. Rewrite the equation in standard form. 6. Solve for the resulting quadratic equation using any of the 4 methods.
Example: Solve the following equations. 1. 6 đ‘„ + đ‘„âˆ’3 4 = 2 24 + đ‘„ đ‘„ − 3 = 8đ‘„ 24 + đ‘„2 − 3đ‘„ = 8đ‘„ 24 + đ‘„2 − 3đ‘„ − 8đ‘„ = 0 24 + đ‘„2 − 11đ‘„ = 0 đ‘„2 − 11đ‘„ + 24 = 0 6 đ‘„ + đ‘„âˆ’3 4 = 2 [4đ‘„] [4đ‘„] 24đ‘„ đ‘„ + 4đ‘„(đ‘„âˆ’3) 4 = 8đ‘„
Example: Solve the following equations. 1. 6 đ‘„ + đ‘„âˆ’3 4 = 2  đ‘„2 − 11đ‘„ + 24 = 0 đ‘„2 − 11đ‘„ + 24 = 0 đ‘„ − 3 đ‘„ − 8 = 0 đ‘„ − 3 = 0 or đ‘„ − 8 = 0 đ‘„ = 3 or đ‘„ = 8
Solving Rational Algebraic Equations Transformable to Quadratic Equations using Cross-Multiplication 1. Cross-multiply the means and the extremes. 2. Apply the inverse operation. 3. Rewrite the equation in standard form. 4. Solve for the resulting quadratic equation using any of the 4 methods. Cross-Multiplication Property: If 𝑎 𝑏 = 𝑐 𝑑 , then 𝑎𝑑 = 𝑏𝑐. then 𝑎𝑑 = 𝑏𝑐.
Example: Solve the following equations. 1. 4đ‘„âˆ’6 2đ‘„âˆ’3 = 7 đ‘„+1 4đ‘„2 − 2đ‘„ − 6 = 14đ‘„ − 21 4đ‘„2 − 2đ‘„ − 14đ‘„ − 6 + 21 = 14đ‘„ − 14đ‘„ − 21 + 21 4đ‘„2 − 16đ‘„ + 15 = 0 4đ‘„2 − 6đ‘„ − 10đ‘„ + 15 = 0 4đ‘„ − 6 2đ‘„ − 3 = 7 đ‘„ + 1 4đ‘„ − 6 đ‘„ + 1 = 7(2đ‘„ − 3) EXTREMES MEANS EXTREMES MEANS
Example: Solve the following equations. 1. 4đ‘„âˆ’6 2đ‘„âˆ’3 = 7 đ‘„+1  4đ‘„2 − 16đ‘„ + 15 = 0 4đ‘„2 − 16đ‘„ + 15 = 0 (4đ‘„2 − 6đ‘„)(−10đ‘„ + 15) = 0 2đ‘„ 2đ‘„ − 3 − 5 2đ‘„ − 3 = 0 2đ‘„ − 3 2đ‘„ − 5 = 0 2đ‘„ − 3 = 0 or 2đ‘„ − 5 = 0 2đ‘„ = 3 or 2đ‘„ = 5 đ‘„ = 3 2 or đ‘„ = 5 2
Solve the following equations. DRILL 1. 10 đ‘„ + 1 = 3đ‘„
Problem Solving involving Quadratic Equations
PROBLEM # 1: The sum of two numbers is 22 and the sum of their squares is 250. Find the numbers. STEP 1: Define the variables. Let đ‘„ = one of the two numbers 22 − đ‘„ = the other number STEP 2: Identify the thought process. one of the two numbers 2 + the other number 2 = 250
PROBLEM # 1: The sum of two numbers is 22 and the sum of their squares is 250. Find the numbers. STEP 3: Formulate the equation. đ‘„2 + (22 − đ‘„)2 = 250 STEP 4: Solve the equation. đ‘„2 + (22 − đ‘„)2 = 250 đ‘„2 + 484 − 44đ‘„ + đ‘„2 = 250 2đ‘„2 − 44đ‘„ + 234 = 0 2đ‘„2 − 44đ‘„ + 234 2 = 0 2
PROBLEM # 1: The sum of two numbers is 22 and the sum of their squares is 250. Find the numbers. STEP 4: Solve the equation. đ‘„2 − 22đ‘„ + 117 = 0 (đ‘„ − 13)(đ‘„ − 9) = 0 đ‘„ = 13 𝑜𝑟 đ‘„ = 9 đŒđ‘“ đ‘„ = 13, 𝑡ℎ𝑒𝑛 22 − 13 = 9. đŒđ‘“ đ‘„ = 9, 𝑡ℎ𝑒𝑛 22 − 9 = 13. The numbers are 9 and 13. To Check: 9 + 13 = 22 92 + 132 = 81 + 169 = 250
Solve the word problem below. DRILL PROBLEM: Two numbers differ by 9. The sum of their squares is 653. Identify the 2 numbers. Let đ‘„ = the smaller number đ‘„ + 9 = the larger number
PROBLEM # 2: If the length of each side of a square is increased by 5 cm, the area is multiplied by 4. What is the length of the original side of the square? STEP 1: Define the variables. Let 𝑠 = one of the two numbers 𝑠2 = the other number STEP 2: Identify the thought process. one of the two numbers + 5 2 = 4 area of the square
PROBLEM # 2: If the length of each side of a square is increased by 5 cm, the area is multiplied by 4. What is the length of the original side of the square? STEP 3: Formulate the equation. (𝑠 + 5)2 = 4(𝑠2 ) STEP 4: Solve the equation. (𝑠 + 5)2 = 4(𝑠2 ) 𝑠2 + 10𝑠 + 25 = 4𝑠2 −4𝑠2 + 𝑠2 + 10𝑠 + 25 = 4𝑠2 − 4𝑠2 0 = 3𝑠2 − 10𝑠 − 25 0 = 3𝑠2 + 5𝑠 − 15𝑠 − 25 0 = (3𝑠2 + 5𝑠)(−15𝑠 − 25)
PROBLEM # 2: If the length of each side of a square is increased by 5 cm, the area is multiplied by 4. What is the length of the original side of the square? STEP 4: Solve the equation. 0 = 3𝑠2 + 5𝑠 −15𝑠 − 25 0 = 𝑠 3𝑠 + 5 − 5 3𝑠 + 5 0 = (3𝑠 + 5)(𝑠 − 5) 3𝑠 + 5 = 0 𝑜𝑟 𝑠 − 5 = 0 𝑠 = − 5 3 𝑜𝑟 𝑠 = 5 Since a side of the square is a physical quantity that cannot assume a negative value, we can only accept 𝑠 = 5. Therefore, the length of the original side of the square is 5 cm.
PROBLEM # 3: A rectangle is 30 cm long and 20 cm wide. A rectangular strip added to one side and another of the same width to the other side result in the doubling of the area. Find the width of the strip. STEP 1: Draw the figure.
PROBLEM # 3: A rectangle is 30 cm long and 20 cm wide. A rectangular strip added to one side and another of the same width to the other side result in the doubling of the area. Find the width of the strip. STEP 3: Formulate the equation. 30 + đ‘€ 20 + đ‘€ = 2(30)(20) STEP 4: Solve the equation. 600 + 50đ‘€ + đ‘€2 = 2(30)(20) đ‘€2 + 50đ‘€ + 600 = 1200 đ‘€2 + 50đ‘€ + 600 − 1200 = 1200 − 1200 Let 𝒘 = width of the strip STEP 2: Define the variable.
STEP 4: Solve the equation. đ‘€2 + 50đ‘€ + 600 − 1200 = 0 đ‘€2 + 50đ‘€ − 600 = 0 đ‘€ + 60 đ‘€ − 10 = 0 đ‘€ + 60 = 0 𝑜𝑟 đ‘€ − 10 = 0 đ‘€ = −60 𝑜𝑟 đ‘€ = 10 Since the width cannot be negative, we can only accept đ‘€ = 10. Therefore, the width of the rectangular strip is 10 cm. PROBLEM # 3: A rectangle is 30 cm long and 20 cm wide. A rectangular strip added to one side and another of the same width to the other side result in the doubling of the area. Find the width of the strip.
Solve the word problem below. DRILL PROBLEM: The perimeter of a rectangle is 60 m and its area is 𝟐𝟐𝟏 𝒎𝟐 . Find the length and the width of the rectangle.

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Quadratic Equations in One Variables.pptx

  • 2. Examples: ï±đ‘„2 + 5đ‘„ + 3 = 0 2đ‘„2 − 4đ‘„ + 9 = 0 ï±đ‘„ 3 + đ‘„2 2 = 5  đ‘„ − 2 đ‘„ + 3 = 0 ï±đ‘„2 − 5 = đ‘„ + 6 ï±đ‘„ đ‘„ − 4 = 1 Quadratic Equation in One Variable A quadratic equation in one variable is of the form đ‘Žđ‘„2 + đ‘đ‘„ + 𝑐 = 0, where 𝑎, 𝑏, and 𝑐 are real number coefficients and 𝑎 ≠ 0. This equation is in the second degree wherein the highest exponent is 2.
  • 3. đ‘„ − 2 đ‘„ + 3 = 0 Rewrite each equation in standard form. Then, identify the value of 𝒂, 𝒃, and 𝒄. 1. đ‘„ − 2 đ‘„ + 3 = 0 F O I L F: đ‘„2 O: 3đ‘„ I: −2đ‘„ L: −6 Quadratic Equation: đ‘„2 − đ‘„ − 6 = 0 Therefore, 𝑎 = 1; 𝑏 = −1; 𝑐 = −6.
  • 4. Rewrite each equation in standard form. Then, identify the value of 𝒂, 𝒃, and 𝒄. 2. đ‘„2 − 5 = đ‘„ + 6 Quadratic Equation: đ‘„2 − đ‘„ + 1 = 0 đ‘„2 − 5 = đ‘„ + 6 đ‘„2 − 5 − đ‘„ = đ‘„ − 6 đ‘„2 − đ‘„ − 5 + 6 = −6 + 6 đ‘„2 − đ‘„ + 1 Therefore, 𝑎 = 1; 𝑏 = −1; 𝑐 = 1.
  • 5. Rewrite each equation in standard form. Then, identify the value of 𝒂, 𝒃, and 𝒄. 3. (2đ‘„ + 1)2 = 5đ‘„ Quadratic Equation: 4đ‘„2 − đ‘„ + 1 = 0 (2đ‘„ + 1)2 = 5đ‘„ 4đ‘„2 + 4đ‘„ + 1 = 5đ‘„ 4đ‘„2 + 4đ‘„ − 5đ‘„ + 1 = 5đ‘„ − 5đ‘„ 4đ‘„2 − đ‘„ + 1 = 0 Therefore, 𝑎 = 4; 𝑏 = −1; 𝑐 = 1.
  • 6. Rewrite each equation in standard form. Then, identify the value of 𝒂, 𝒃, and 𝒄. DRILL 1. (đ‘„ + 5)2 = 3 2. đ‘„ đ‘„ + 2 = 5đ‘„2 − 4 3. 2đ‘„ + 1 đ‘„ − 3 = đ‘„ + 1 4. đ‘„ − 4 đ‘„ − 6 = 2 5. 3 4 đ‘„2 − 2 3 đ‘„ = 3 5
  • 8. Nature of the Roots If a, b, and c are real numbers and đ· = 𝑏2 − 4𝑎𝑐, then the roots of the quadratic equation đ‘Žđ‘„2 + đ‘đ‘„ + 𝑐 = 0 are: I. real and unequal, if D > 0; II. real and equal, if D = 0; III. imaginary and unequal, if D < 0; Moreover, the roots are: IV. rational and unequal, if D is a perfect square; V. irrational and unequal, if D is not a perfect square. Discriminant (𝒃𝟐 − 𝟒𝒂𝒄) The discriminant provided information regarding the nature of the roots of the quadratic equation.
  • 9. Determine the nature of the roots of each equation. 1. đ‘„2 − 6đ‘„ + 9 = 0 đ‘„2 − 6đ‘„ + 9 = 0; a = 1, b = -6, and c = 9 đ· = 𝑏2 − 4𝑎𝑐 đ· = (−6)2 −4(1)(9) đ· = 36 − 36 đ· = 0 Therefore, the roots of 𝒙𝟐 − 𝟔𝒙 + 𝟗 = 𝟎 are real, equal, and rational.
  • 10. Determine the nature of the roots of each equation. 2. 2đ‘„2 − 4đ‘„ + 5 = 0 2đ‘„2 − 4đ‘„ + 5 = 0; a = 2, b = -4, and c = 5 đ· = 𝑏2 − 4𝑎𝑐 đ· = (−4)2 −4(2)(5) đ· = 16 − 40 đ· = −24 Therefore, the roots of 𝟐𝒙𝟐 − 𝟒𝒙 + 𝟓 = 𝟎 are imaginary and unequal.
  • 11. Determine the nature of the roots of each equation. DRILL 1. đ‘„2 + 10đ‘„ + 22 = 0 2. 4đ‘„2 + 2đ‘„ + 4 = 0 3. 2đ‘„2 − 4đ‘„ − 4 = 0 4. đ‘„2 − 10đ‘„ + 22 = 0 5. 3 4 đ‘„2 − 2 3 đ‘„ − 3 5 = 0
  • 12. Sum and Product of the Roots Sum of The Roots = − 𝑏 𝑎 ; Product of the Roots = 𝑐 𝑎 Find the sum and product of the roots of the following equations. 1. đ‘„2 − 6đ‘„ + 9 = 0 đ‘„2 − 6đ‘„ + 9 = 0; a = 1, b = -6, and c = 9 𝑆𝑈𝑀 = −𝑏 𝑎 đ‘ƒđ‘…đ‘‚đ·đ‘ˆđ¶đ‘‡ = 𝑐 𝑎 = −(−6) 1 = 6 = 6 1 = 9 1 = 9
  • 13. Find the sum and product of the roots of the following equations. 2. 2đ‘„2 − 4đ‘„ + 5 = 0 2đ‘„2 − 4đ‘„ + 5 = 0; a = 2, b = -4, and c = 5 𝑆𝑈𝑀 = −𝑏 𝑎 đ‘ƒđ‘…đ‘‚đ·đ‘ˆđ¶đ‘‡ = 𝑐 𝑎 = 2 = 4 2 = −(−4) 2 = 2.5 = 5 2
  • 14. Determine the sum and products of the roots of each equation. DRILL 1. đ‘„2 + 10đ‘„ + 22 = 0 2. 4đ‘„2 + 2đ‘„ + 4 = 0 3. 2đ‘„2 − 4đ‘„ − 4 = 0 4. đ‘„2 − 10đ‘„ + 22 = 0 5. 3 4 đ‘„2 − 2 3 đ‘„ − 3 5 = 0
  • 15. Solving Quadratic Equations by: Extracting Square Roots Factoring Completing The Square Using The Quadratic Formula
  • 16. No. of Solutions in a Quadratic E quation  Every quadratic equation in one variable has exactly two solutions or roots.
  • 17. Square Root Property  If đ‘„2 = 𝑛, then đ‘„ = 𝑛 or đ‘„ = − 𝑛.
  • 18. ‱ If 𝑏 = 0, the equation is a pure or incomplete quadratic equation. ‱ Example: đ‘„2 − 4 = 0 and 3đ‘„2 + 6 = 0 ‱ If 𝑏 ≠ 0, the equation is a complete quadratic equation. ‱ Example: đ‘„2 − 4đ‘„ + 1 = 0 and 3đ‘„2 + 2đ‘„ − 1 = 0 A. Extracting Square Roots Step 1: Express the quadratic equation in standard form. Step 2: Factor the quadratic expression. Step 3: Apply the zero-product property and set each variable factor equal to 0. Step 4: Solve the resulting linear equations.
  • 19. Solve the quadratic equation by extracting the square root. 1. đ‘„2 − 11 = 0 đ‘„2 − 11 = 0 đ‘„2 = 11 đ‘„ = ± 11 To Check: If đ‘„ = 11, If đ‘„ = − 11 ( 11)2 −11 = 0 − 11 2 − 11 = 0 11 − 11 = 0 11 − 11 = 0 0 = 0 True 0 = 0 True
  • 20. Solve the quadratic equation by extracting the square root. 2. 2đ‘„2 + 3 = 39 2đ‘„2 + 3 = 39 2đ‘„2 = 36 đ‘„2 = 18 đ‘„ = ± 18 đ‘„ = ± (9)(2) đ‘„ = ± 9 2 đ‘„ = ±3 2 To Check: If đ‘„ = 2 3, If đ‘„ = −3 2 2(3 2)2 + 3 = 39 −3 2 2 + 3 = 39 2(9)(2) + 3 = 39 2 9 2 + 3 = 39 36 + 3 = 39 36 + 3 = 39 39 = 39 True 39 = 39 True
  • 21. Solve the quadratic equation by extracting the square root. 3. (đ‘„ + 4)2 = 1 (đ‘„ + 4)2= 1 đ‘„ + 4 2 = 1 đ‘„ + 4 = ±1 To Check: If đ‘„ = −3, If đ‘„ = −5 (−3 + 4)2 = 1 −5 + 4 2 = 1 (1)2= 1 (−1)2= 1 1 = 1 True 1 = 1 True For đ‘„ + 4 = 1, đ‘„ = −3 For đ‘„ + 4 = −1, đ‘„ = −5
  • 22. Solve the following quadratic equations using the Square Root Property. DRILL 1. 6đ‘„2 − 24 = 0 2. 2đ‘„2 − 34 = 0 3. (đ‘„ + 2)2 = 9 4. (4đ‘„ + 1)2 = 196
  • 23. Zero Property of Equality  The product đŽđ” = 0, if and only if 𝐮 = 0 or đ” = 0.
  • 24. B. Factoring Step 1: Clear all fractions (if any) and write the given equation in the form đ‘Žđ‘„2 + đ‘đ‘„ + 𝑐 = 0. Step 2: Factor the product of the coefficient of đ‘„ and the constant term of the given quadratic equation. Step 3: Express the coefficient of the middle term as the sum or difference of the factors obtained in Step 2. Step 4: Use zero product property to put each linear factor equal to 0. Step 5: Solve the resulting linear equations.
  • 25. Solve the quadratic equation by factoring. 1. đ‘„2 − 9đ‘„ + 18 = 0 đ‘„2 − 9đ‘„ + 18 = 0 đ‘„ − 3 đ‘„ − 6 = 0 đ‘„ − 3 = 0 đ‘„ − 6 = 0 đ‘„ = 3 đ‘„ = 6 TO CHECK: đ‘„ = 3 đ‘„2 − 9đ‘„ + 18 = 0 32 − 9 3 + 18 = 0 9 − 27 + 18 = 0 9 − 9 = 0 0 = 0 True TO CHECK: đ‘„ = 6 đ‘„2 − 9đ‘„ + 18 = 0 62 − 9 6 + 18 = 0 36 − 54 + 18 = 0 36 − 36 0 = 0 True
  • 26. Solve the quadratic equation by factoring. 2. đ‘„2 + 12 = 7đ‘„ đ‘„2 + 12 = 7đ‘„ đ‘„2 − 7đ‘„ + 12 = 0 đ‘„ − 3 đ‘„ − 4 = 0 đ‘„ − 3 = 0 đ‘„ − 4 = 0 đ‘„ = 3 đ‘„ = 4 TO CHECK: đ‘„ = 3 đ‘„2 + 12 = 7đ‘„ 32 + 12 = 7(3) 9 + 12 = 21 21 = 21 True TO CHECK: đ‘„ = 4 đ‘„2 + 12 = 7đ‘„ 42 + 12 = 7(4) 16 + 12 = 28 28 = 28 True
  • 27. Solve the following quadratic equations by factoring. DRILL 1. đ‘„2 − 5đ‘„ + 6 = 0 2. đ‘„2 − 3đ‘„ − 18 = 0 3. 2đ‘„2 + 3đ‘„ − 2 = 0 4. 2đ‘„2 + 7đ‘„ − 4 = 0
  • 28. C. Completing the Square Step 1: Rewrite the equation in the form đ‘Žđ‘„2 + đ‘đ‘„ + 𝑐 = 0. Step 2: Add to both sides the term needed to complete the square. Step 3: Factor the perfect square trinomial. Step 4: Solve the resulting equation by using the square root property.. ‱ To make an expression of the form đ‘„2 + đ‘đ‘„ a perfect square trinomial, add 𝑏 2 2 to it.
  • 29. Solve the quadratic equation by completing the square. 1. đ‘„2 + 6đ‘„ + 8 = 0 đ‘„2 + 6đ‘„ + 8 = 0 đ‘„2 + 6đ‘„ = −8 đ‘„2 + 6đ‘„ + 6 2 2 = −8 đ‘„2 + 6đ‘„ + (3)2= −8 đ‘„2 + 6đ‘„ + 9 = −8 + 9 đ‘„ + 3 2 = 1 (đ‘„ + 3)2= ± 1
  • 30. Solve the quadratic equation by completing the square. 1. đ‘„2 + 6đ‘„ + 8 = 0 đ‘„ + 3 = ±1 đ‘„ = −3 ± 1 đ‘„ = −3 + 1 = −2 or đ‘„ = −3 − 1 = −4 TO CHECK: đ‘„ = −2 đ‘„2 + 6đ‘„ + 8 = 0 −22 + 6 −2 + 8 = 0 4 − 12 + 8 = 0 4 − 4 = 0 0 = 0 True TO CHECK: đ‘„ = −4 đ‘„2 + 6đ‘„ + 8 = 0 −42 + 6 −4 + 8 = 0 16 − 24 + 8 = 0 16 − 16 = 0 0 = 0 True SOLUTIONS: −2 & − 4
  • 31. Solve the quadratic equation by completing the square. 2. đ‘„2 − 2đ‘„ = 1 đ‘„2 − 2đ‘„ = 1 đ‘„2 − 2đ‘„ + 2 2 2 = 1 đ‘„2 − 2đ‘„ + 1 2 = 1 đ‘„2 − 2đ‘„ + 1 = 1 + 1 (đ‘„ − 1)2= 2 (đ‘„ − 1)2= ± 2 đ‘„ = 1 + 2 or đ‘„ = 1 − 2 SOLUTIONS:1 + 2 & 1 − 2
  • 32. Solve the quadratic equation by completing the square. 3. 4đ‘„2 − 5đ‘„ − 6 = 0 4đ‘„2 − 5đ‘„ − 6 = 0 4đ‘„2 − 5đ‘„ = 6 4 (đ‘„2 − 5 4 đ‘„) = 6 4 (đ‘„2 − 5 4 đ‘„) 4 = 6 4 đ‘„2 − 5 4 đ‘„ = 3 2
  • 33. Solve the quadratic equation by completing the square. 3. 4đ‘„2 − 5đ‘„ − 6 = 0 Complete the Square: 5 4 2 2 = 5 8 2 = 25 64 đ‘„2 − 5 4 đ‘„ + 25 64 = 3 2 + 25 64 Add the fractions on the right side. 3 2 + 25 64 = 96 64 + 25 64 = 121 64
  • 34. Solve the quadratic equation by completing the square. 3. 4đ‘„2 − 5đ‘„ − 6 = 0 đ‘„2 − 5 4 đ‘„ + 25 64 = 121 64 đ‘„ − 5 8 2 = 121 64 đ‘„ − 5 8 2 = ± 121 64
  • 35. Solve the quadratic equation by completing the square. 3. 4đ‘„2 − 5đ‘„ − 6 = 0 đ‘„ − 5 8 = ± 11 8 đ‘„ = 5 8 ± 11 8 đ‘„ = 5 8 + 11 8 = 16 8 = 2 đ‘„ = 5 8 − 11 8 = − 6 8 = − 3 4 SOLUTIONS: 2 & − 3 4
  • 36. Solve the following quadratic equations by completing the square. DRILL 1. đ‘„2 − 2đ‘„ − 8 = 0 2. đ‘„2 + 7đ‘„ − 18 = 0 3. 3đ‘„2 = 2 − đ‘„ 4. đ‘„2 = 9
  • 37. D. Using the Quadratic Formula đ‘„ = −𝑏 ± 𝑏2 − 4𝑎𝑐 2𝑎 đ‘„ = −𝑏 ± 𝑏2 − 4𝑎𝑐 2𝑎 đ‘„ = −9 ± 92 − 4(1)(14) 2(1) Solve the quadratic equation using the quadratic formula. 1. đ‘„2 + 9đ‘„ + 14 = 0
  • 38. Solve the quadratic equation using the quadratic formula. 1. đ‘„2 + 9đ‘„ + 14 = 0 đ‘„ = −9 ± 81 − 56 2 đ‘„ = −9 ± 25 2 đ‘„ = −9 ± 5 2
  • 39. Solve the quadratic equation using the quadratic formula. 1. đ‘„2 + 9đ‘„ + 14 = 0 đ‘„ = −9 ± 5 2 đ‘„ = −9+5 2 or đ‘„ = −9−5 2 đ‘„ = −4 2 or đ‘„ = −14 2 đ‘„ = −2 or đ‘„ = −7
  • 40. Solve the following quadratic equations using the quadratic formula. DRILL 1. đ‘„2 − 2đ‘„ + 1 = 0 2. đ‘„2 − 6đ‘„ + 9 = 0 3. đ‘„2 − 5đ‘„ − 5 = 0 4. 3đ‘„2 − 5đ‘„ − 2 = 0
  • 41. Equations in Quadratic Form (𝑎𝑓2 + 𝑏𝑓 + 𝑐 = 0)
  • 42. Equations in Quadratic Form An equation in quadratic form is any equation of the form 𝑎𝑓2 + 𝑏𝑓 + 𝑐 = 0, where 𝑓 is any algebraic expression. These equations can be solved using simple substitution. Example: Solve the following equations. 1. 9đ‘„4 − 52đ‘„2 + 64 = 0 In 9(đ‘„2 )2 −52 đ‘„2 + 64 = 0, let 𝑓 = đ‘„2 . 9𝑓2 − 52𝑓 + 64 = 0 𝑓 − 4 9𝑓 − 16 = 0 𝑓 − 4 = 0 or 9𝑓 − 16 = 0 𝑓 = 4 or 𝑓 = 16 9
  • 43. Example: Solve the following equations. 1. 9đ‘„4 − 52đ‘„2 + 64 = 0 Let us solve for đ‘„. đ‘„2 = 𝑓 = 4 or đ‘„2 = 𝑓 = 16 9 đ‘„2 = 4 or đ‘„2 = 16 9 đ‘„ = ± 4 or đ‘„ = ± 16 9 đ‘„ = ±2 or đ‘„ = ± 4 3 Solution Set: −2, − 4 3 , 4 3 , 2
  • 44. Example: Solve the following equations. 2. (2đ‘„ − 1)2 − 6 2đ‘„ − 1 = −8 In(2đ‘„ − 1)2−6 2đ‘„ − 1 = −8, let 𝑔 = 2đ‘„ − 1. 𝑔2 − 6𝑔 = −8 𝑔2 − 6𝑔 + 8 = 0 𝑔 − 4 𝑔 − 2 = 0 𝑔 − 4 = 0 or 𝑔 − 2 = 0 𝑔 = 4 or 𝑔 = 2
  • 45. Example: Solve the following equations. 2. (2đ‘„ − 1)2 − 6 2đ‘„ − 1 = −8 Let us solve for đ‘„. 2đ‘„ − 1 = 𝑔 = 4 or 2đ‘„ − 1 = 𝑔 = 2 2đ‘„ − 1 = 4 or 2đ‘„ − 1 = 2 2đ‘„ = 5 or 2đ‘„ = 3 đ‘„ = 5 2 or đ‘„ = 3 2 Solution Set: 5 2 , 3 2
  • 46. Solve for 𝒙 in each equation. DRILL 1. đ‘„4 + 4 = 5đ‘„2 2. đ‘„4 − 17đ‘„2 + 16 = 0
  • 48. Rational Algebraic Equations Rational algebraic equations are equations that contain one or more rational algebraic expressions (algebraic expressions in the form 𝑛 𝑑 , where 𝑛 and 𝑑 are polynomials and 𝑑 ≠ 0). These equations can be reduced to polynomial equations by clearing off fractions. Examples: 5đ‘„ 6 − 1 2 = 1 3 đ‘„ 2 + 3đ‘„ = 7 4đ‘„âˆ’6 2đ‘„âˆ’3 = 7 đ‘„+1
  • 49. But, take note, not all rational algebraic equations can be transformed into quadratic equations. Rational Algebraic Equations Transformable to Quadratic Equations 1. It has to be a rational equation. 2. There should only be one variable. 3. If the expressions are linear, then there should be a variable on both numerator and denominator. Example: 4đ‘„âˆ’6 2đ‘„âˆ’3 4. If the denominators are all constant, then the numerator should have at least one variable that has a power of 2. Example: đ‘„2+1 7
  • 50. Solving Rational Algebraic Equations Transformable to Quadratic Equations by Finding the LCD 1. Multiply both sides of the equation by the LCD. 2. Apply the distributive law. 3. Combine like terms. 4. Apply the inverse operation. 5. Rewrite the equation in standard form. 6. Solve for the resulting quadratic equation using any of the 4 methods.
  • 51. Example: Solve the following equations. 1. 6 đ‘„ + đ‘„âˆ’3 4 = 2 24 + đ‘„ đ‘„ − 3 = 8đ‘„ 24 + đ‘„2 − 3đ‘„ = 8đ‘„ 24 + đ‘„2 − 3đ‘„ − 8đ‘„ = 0 24 + đ‘„2 − 11đ‘„ = 0 đ‘„2 − 11đ‘„ + 24 = 0 6 đ‘„ + đ‘„âˆ’3 4 = 2 [4đ‘„] [4đ‘„] 24đ‘„ đ‘„ + 4đ‘„(đ‘„âˆ’3) 4 = 8đ‘„
  • 52. Example: Solve the following equations. 1. 6 đ‘„ + đ‘„âˆ’3 4 = 2  đ‘„2 − 11đ‘„ + 24 = 0 đ‘„2 − 11đ‘„ + 24 = 0 đ‘„ − 3 đ‘„ − 8 = 0 đ‘„ − 3 = 0 or đ‘„ − 8 = 0 đ‘„ = 3 or đ‘„ = 8
  • 53. Solving Rational Algebraic Equations Transformable to Quadratic Equations using Cross-Multiplication 1. Cross-multiply the means and the extremes. 2. Apply the inverse operation. 3. Rewrite the equation in standard form. 4. Solve for the resulting quadratic equation using any of the 4 methods. Cross-Multiplication Property: If 𝑎 𝑏 = 𝑐 𝑑 , then 𝑎𝑑 = 𝑏𝑐. then 𝑎𝑑 = 𝑏𝑐.
  • 54. Example: Solve the following equations. 1. 4đ‘„âˆ’6 2đ‘„âˆ’3 = 7 đ‘„+1 4đ‘„2 − 2đ‘„ − 6 = 14đ‘„ − 21 4đ‘„2 − 2đ‘„ − 14đ‘„ − 6 + 21 = 14đ‘„ − 14đ‘„ − 21 + 21 4đ‘„2 − 16đ‘„ + 15 = 0 4đ‘„2 − 6đ‘„ − 10đ‘„ + 15 = 0 4đ‘„ − 6 2đ‘„ − 3 = 7 đ‘„ + 1 4đ‘„ − 6 đ‘„ + 1 = 7(2đ‘„ − 3) EXTREMES MEANS EXTREMES MEANS
  • 55. Example: Solve the following equations. 1. 4đ‘„âˆ’6 2đ‘„âˆ’3 = 7 đ‘„+1  4đ‘„2 − 16đ‘„ + 15 = 0 4đ‘„2 − 16đ‘„ + 15 = 0 (4đ‘„2 − 6đ‘„)(−10đ‘„ + 15) = 0 2đ‘„ 2đ‘„ − 3 − 5 2đ‘„ − 3 = 0 2đ‘„ − 3 2đ‘„ − 5 = 0 2đ‘„ − 3 = 0 or 2đ‘„ − 5 = 0 2đ‘„ = 3 or 2đ‘„ = 5 đ‘„ = 3 2 or đ‘„ = 5 2
  • 56. Solve the following equations. DRILL 1. 10 đ‘„ + 1 = 3đ‘„
  • 58. PROBLEM # 1: The sum of two numbers is 22 and the sum of their squares is 250. Find the numbers. STEP 1: Define the variables. Let đ‘„ = one of the two numbers 22 − đ‘„ = the other number STEP 2: Identify the thought process. one of the two numbers 2 + the other number 2 = 250
  • 59. PROBLEM # 1: The sum of two numbers is 22 and the sum of their squares is 250. Find the numbers. STEP 3: Formulate the equation. đ‘„2 + (22 − đ‘„)2 = 250 STEP 4: Solve the equation. đ‘„2 + (22 − đ‘„)2 = 250 đ‘„2 + 484 − 44đ‘„ + đ‘„2 = 250 2đ‘„2 − 44đ‘„ + 234 = 0 2đ‘„2 − 44đ‘„ + 234 2 = 0 2
  • 60. PROBLEM # 1: The sum of two numbers is 22 and the sum of their squares is 250. Find the numbers. STEP 4: Solve the equation. đ‘„2 − 22đ‘„ + 117 = 0 (đ‘„ − 13)(đ‘„ − 9) = 0 đ‘„ = 13 𝑜𝑟 đ‘„ = 9 đŒđ‘“ đ‘„ = 13, 𝑡ℎ𝑒𝑛 22 − 13 = 9. đŒđ‘“ đ‘„ = 9, 𝑡ℎ𝑒𝑛 22 − 9 = 13. The numbers are 9 and 13. To Check: 9 + 13 = 22 92 + 132 = 81 + 169 = 250
  • 61. Solve the word problem below. DRILL PROBLEM: Two numbers differ by 9. The sum of their squares is 653. Identify the 2 numbers. Let đ‘„ = the smaller number đ‘„ + 9 = the larger number
  • 62. PROBLEM # 2: If the length of each side of a square is increased by 5 cm, the area is multiplied by 4. What is the length of the original side of the square? STEP 1: Define the variables. Let 𝑠 = one of the two numbers 𝑠2 = the other number STEP 2: Identify the thought process. one of the two numbers + 5 2 = 4 area of the square
  • 63. PROBLEM # 2: If the length of each side of a square is increased by 5 cm, the area is multiplied by 4. What is the length of the original side of the square? STEP 3: Formulate the equation. (𝑠 + 5)2 = 4(𝑠2 ) STEP 4: Solve the equation. (𝑠 + 5)2 = 4(𝑠2 ) 𝑠2 + 10𝑠 + 25 = 4𝑠2 −4𝑠2 + 𝑠2 + 10𝑠 + 25 = 4𝑠2 − 4𝑠2 0 = 3𝑠2 − 10𝑠 − 25 0 = 3𝑠2 + 5𝑠 − 15𝑠 − 25 0 = (3𝑠2 + 5𝑠)(−15𝑠 − 25)
  • 64. PROBLEM # 2: If the length of each side of a square is increased by 5 cm, the area is multiplied by 4. What is the length of the original side of the square? STEP 4: Solve the equation. 0 = 3𝑠2 + 5𝑠 −15𝑠 − 25 0 = 𝑠 3𝑠 + 5 − 5 3𝑠 + 5 0 = (3𝑠 + 5)(𝑠 − 5) 3𝑠 + 5 = 0 𝑜𝑟 𝑠 − 5 = 0 𝑠 = − 5 3 𝑜𝑟 𝑠 = 5 Since a side of the square is a physical quantity that cannot assume a negative value, we can only accept 𝑠 = 5. Therefore, the length of the original side of the square is 5 cm.
  • 65. PROBLEM # 3: A rectangle is 30 cm long and 20 cm wide. A rectangular strip added to one side and another of the same width to the other side result in the doubling of the area. Find the width of the strip. STEP 1: Draw the figure.
  • 66. PROBLEM # 3: A rectangle is 30 cm long and 20 cm wide. A rectangular strip added to one side and another of the same width to the other side result in the doubling of the area. Find the width of the strip. STEP 3: Formulate the equation. 30 + đ‘€ 20 + đ‘€ = 2(30)(20) STEP 4: Solve the equation. 600 + 50đ‘€ + đ‘€2 = 2(30)(20) đ‘€2 + 50đ‘€ + 600 = 1200 đ‘€2 + 50đ‘€ + 600 − 1200 = 1200 − 1200 Let 𝒘 = width of the strip STEP 2: Define the variable.
  • 67. STEP 4: Solve the equation. đ‘€2 + 50đ‘€ + 600 − 1200 = 0 đ‘€2 + 50đ‘€ − 600 = 0 đ‘€ + 60 đ‘€ − 10 = 0 đ‘€ + 60 = 0 𝑜𝑟 đ‘€ − 10 = 0 đ‘€ = −60 𝑜𝑟 đ‘€ = 10 Since the width cannot be negative, we can only accept đ‘€ = 10. Therefore, the width of the rectangular strip is 10 cm. PROBLEM # 3: A rectangle is 30 cm long and 20 cm wide. A rectangular strip added to one side and another of the same width to the other side result in the doubling of the area. Find the width of the strip.
  • 68. Solve the word problem below. DRILL PROBLEM: The perimeter of a rectangle is 60 m and its area is 𝟐𝟐𝟏 𝒎𝟐 . Find the length and the width of the rectangle.