Solving Quadratic Equations by Completing the Square
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The document outlines methods for solving quadratic equations by completing the square, detailing steps to express trinomials as perfect square binomials. It provides examples of how to manipulate and solve specific quadratic equations using this technique. The instructions include adding necessary constants to both sides of the equation and extracting square roots to find solutions.
Solving Quadratic Equations by Completing the Square
- 1. Grade 9 – Mathematics Quarter I SOLVING QUADRATIC EQUATIONS BY COMPLETING THE SQUARE
- 2. OBJECTIVES: •Express square of trinomials as a square of binomials; and •Solve quadratic equations by completing the square.
- 3. COMPLETING THE SQUARE To complete the square of the expression 𝑥2 + 𝑏𝑥, add the square of half the coefficient of 𝑥 to make 𝒙 𝟐 +𝒃𝒙 + 𝒃 𝟐 𝟐 .
- 4. Complete the square. 𝒙 𝟐 − 𝟖𝒙 + ______ 𝒙 𝟐 + +𝒃𝒙 + 𝒃 𝟐 𝟐 − 8 2 = −4 (−4)2 = 16 𝒙 𝟐 − 𝟖𝒙 + 𝟏𝟔 (𝒙 − 𝟒) 𝟐
- 5. Complete the square. 𝒙 𝟐 + 𝟑𝒙 + ______ 𝒙 𝟐 + +𝒃𝒙 + 𝒃 𝟐 𝟐 3 2 = 3 2 3 2 2 = 9 4 𝒙 𝟐 + 𝟑𝒙 + 𝟗 𝟒 𝑥 + 3 2 2
- 6. Express each of the following perfect square trinomials as a square of a binomial. 1. 𝑥2 + 4𝑥 + 4 (𝑥 + 2)2 2. 𝑥2 + 12𝑥 + 36 (𝑥 + 6)2 3. 𝑥2 − 10𝑥 + 25 (𝑥 − 5)2 4. 𝑥2 − 5𝑥 + 25 4 (𝑥 − 5 2 )2
- 7. 𝑥2 + 10𝑥 = 3 𝑥2 + 10𝑥 + ______ 10 2 = 5 (5)2 = 25 𝑥2 + 10𝑥 +25 (𝒙 + 𝟓) 𝟐 𝑥2 + 10𝑥 + 25 = 3 + 25 Add 25 to both sides of the equation Express the PST on the left side of the equation as square of binomial. (𝑥 + 5)2 = 28 Solve the resulting QE by extracting the square root. 𝑥 + 5 = ± 28 Solve each equation. 𝑥 + 5 = 2 7 𝑥 + 5 = −2 7 𝑥 = −5 + 2 7 𝑥 = −5 − 2 7 𝑥 + 5 = ±2 7 28 = 4 ∙ 7 = 2 7
- 8. 𝑥2 − 3𝑥 = 1𝑥2 − 3𝑥 + ______ − 3 2 − 3 2 2 = 9 4 𝑥2 − 3𝑥 + 9 4 𝑥2 − 3𝑥 + 9 4 = 1 + 9 4 Add 9 4 to both sides of the equation Express the PST on the left side of the equation as square of binomial. x − 3 2 2 = 13 4 Solve the resulting QE by extracting the square root. 𝒙 − 𝟑 𝟐 𝟐 𝑥 − 3 2 = ± 13 4 𝑥 − 3 2 = ± 13 2
- 9. Solve each equation. 𝑥 − 3 2 = 13 2 𝑥 − 3 2 = − 13 2 𝑥 = 3 2 + 13 2 𝑥 = 3 2 − 13 2 𝒙 = 𝟑 + 𝟏𝟑 𝟐 𝒙 = 𝟑 − 𝟏𝟑 𝟐 𝑥 − 3 2 = ± 13 2
- 10. 𝑥2 + 5𝑥 + 4 = 0 𝑥2 + 5𝑥 + _____ 5 2 5 2 2 = 25 4 𝑥2 + 5𝑥 + 25 4 𝑥2 + 5𝑥 + 25 4 = −4 + 25 4 Add 25 4 to both sides of the equation Express the PST on the left side of the equation as square of binomial. x + 5 2 2 = 9 4 Solve the resulting QE by extracting the square root. 𝒙 + 𝟓 𝟐 𝟐 𝑥 + 5 2 = ± 9 4 𝑥 + 5 2 = ± 3 2 𝑥2 + 5𝑥 = −4
- 11. Solve each equation. 𝑥 + 5 2 = 3 2 𝑥 + 5 2 = − 3 2 𝑥 = − 5 2 + 3 2 𝒙 = −𝟏 𝑥 + 5 2 = ± 3 2 𝑥 = − 2 2 𝑥 = − 5 2 − 3 2 𝒙 = −𝟒 𝑥 = − 8 2
- 12. 2𝑥2 − 3𝑥 − 9 = 0 Add 9 16 to both sides of the equation Express the PST on the left side of the equation as square of binomial. Solve the resulting QE by extracting the square root. 2𝑥2 − 3𝑥 = 9 Divide the terms by 2. 𝑥2 − 3𝑥 2 = 9 2 − 3 2 2 = − 3 2 ∙ 1 2 = − 3 4 − 3 4 2 = 9 16 𝑥2 − 3𝑥 2 + 9 16 𝒙 − 𝟑 𝟒 𝟐 𝑥2 − 3𝑥 2 + 9 16 = 9 2 + 9 16 𝑥 − 3 4 2 = 81 16 𝑥 − 3 4 = ± 81 16 𝑥 − 3 4 = ± 9 4
- 13. Solve each equation. 𝑥 = 3 4 + 9 4 𝑥 − 3 4 = 9 4 𝑥 − 3 4 = ± 9 4 𝑥 − 3 4 = − 9 4 𝑥 = 3 4 − 9 4 𝑥 = −6 4𝑥 = 12 4 𝒙 = 𝟑 𝒙 = − 𝟑 𝟐