MIT Math Syllabus 10-3 Lesson 7: Quadratic equations

Definition of a Quadratic Equation
A quadratic equation in x is an equation that can be written in
the standard quadratic form
ax2 + bx + c = 0
where a, b and c are real numbers and a  0.
Several methods can be used to solve a quadratic equation.

SOLVING QUADRATIC EQUATIONS
There are four algebraic methods of solving quadratic
equation in one variable, namely:
•solution by factoring
•solution by extraction of roots or taking square roots
•solution by completing the square
•solution by quadratic formula

SOLVING QUADRATIC EQUATIONS BY FACTORING

For instance, if you can factor ax2 + bx + c into linear factors, then
ax2 + bx + c = 0 can be solved by applying
the following property.
The Zero Product Principle
If A and B are algebraic expressions such that AB = 0, then A = 0
or B = 0.
The zero product principle states that if the product of two
factors is zero, then at least one of the factors must be zero.
In Example 1, the zero product principle is used to solve a
quadratic equation.

SOLVE BY FACTORING
Solve each quadratic equation by factoring.
a. x2 + 2x – 15 = 0 b. 2x2 – 5x = 12
Solution:
a. x2 + 2x – 15 = 0
(x – 3)(x + 5) = 0
x – 3 = 0 or x + 5 = 0
x = 3 x = –5
A check shows that 3 and –5 are both solutions of
x2 + 2x – 15 = 0.
Factor.
Set each factor equal to zero.
Solve each linear equation.
EXAMPLE

SOLUTION
b. 2x2 – 5x = 12
2x2 – 5x – 12 = 0
(x – 4)(2x + 3) = 0
x – 4 = 0 or 2x + 3 = 0
x = 4 2x = –3
x =
A check shows that 4 and are both solutions of
2x2 – 5x = 12.
cont’d
Write in standard quadratic form.
Factor.
Set each factor equal to zero.

Some quadratic equations have a solution that is called a double
root. For instance, consider x2 – 8x + 16 = 0.
Solving this equation by factoring, we have
x2 – 8x + 16 = 0
(x – 4)(x – 4) = 0
x – 4 = 0 or x – 4 = 0
x = 4 x = 4
Factor.
Set each factor equal to zero

The only solution of x2 – 8x + 16 = 0 is 4. In this situation,
the single solution 4 is called a double solution or double
root because it was produced by solving the two identical
equations x – 4 = 0, both of which have 4 as a solution.

EXERCISES: SOLVE EACH QUADRATIC EQUATIONS
xx
xxxx
x
x
xx
xxxx
xxxx
xxxx
55.11
0644.100103.5
34
2
.9093025.4
13)13)(14(.801024.3
3)52(.7.0168.2
15812.6030.1
2
22
2
2
2
22







SOLVING QUADRATIC EQUATIONS BY TAKING SQUARE ROOTS

Recall that This principle can be used to solve some
quadratic equations by taking the square root of each side of the
equation.
In the following example, we use this idea to solve x2 = 25.
x2 = 25
=
| x | = 5
x = –5 or x = 5
The solutions are –5 and 5.
Take the square root of each side.
Use the fact that = |x| and
= 5.
Solve the absolute value equation.

We will refer to the preceding method of solving a quadratic
equation as the square root procedure.
The Square Root Procedure
If x2 = c, then x = or x = , which can also be written as
x = .
Example
If x2 = 9, then x = = 3 or x = = –3.
This can be written as x = 3.

If x2 = 7, then x = or x = .
This can be written as x =  .
If x2 = –4, then x = = 2i or x = = –2i.
This can be written as x = 2i.

SOLVE BY USING THE SQUARE ROOT PROCEDURE
Use the square root procedure to solve each equation.
a. 3x2 + 12 = 0 b. (x + 1)2 = 48
Solution:
a. 3x2 + 12 = 0
3x2 = –12
x2 = –4
x =
Solve for x2.
Take the square root of each side of the
equation and insert a plus-or-minus sign
in front of the radical.
EXAMPLE

SOLUTION
x = –2i or x = 2i
The solutions are –2i and 2i.
b. (x + 1)2 = 48
cont’d
Take the square root of each side of the
equation and insert a plus-or-minus sign
in front of the radical.
Simplify.

The solutions are and .
cont’d
SOLUTION

0582.5
01255.4
0)7(2y8.09.3
49)3(7.17)3(.2
202076.04.1
2
2
22
22
22





w
x
x
xx
xx

SOLVING QUADRATIC EQUATIONS BY COMPLETING THE
SQUARE

SQUARE
Consider two binomial squares and their perfect-square
trinomial products.
In each of the preceding perfect-square trinomials, the
coefficient of x2 is 1 and the constant term is the square of half
the coefficient of the x term.
x2 + 10x + 25,

x2 – 6x + 9,
Adding to a binomial of the form x2 + bx the constant term that
makes the binomial a perfect-square trinomial is called
completing the square.
For example, to complete the square of x2 + 8x, add
to produce the perfect-square trinomial x2 + 8x + 16.
SQUARE

Completing the square is a powerful procedure that can be used
to solve any quadratic equation.
For instance, to solve x2 – 6x + 13 = 0, first isolate the variable
terms on one side of the equation and the constant term on the
other side.
x2 – 6x = –13
x2 – 6x + 9 = –13 + 9
Subtract 13 from each side of
the equation.
Complete the square by adding
to each side of the
equation.
SQUARE

(x – 3)2 = –4
x – 3 =
x – 3 = 2i
x = 3  2i
The solutions of x2 – 6x + 13 = 0 are 3 – 2i and 3 + 2i.
You can check these solutions by substituting each solution into
the original equation.
Factor and solve by the square
root procedure.
SQUARE

For instance, the following check shows that 3 – 2i does satisfy
the original equation.
x2 – 6x + 13 = 0
(3 – 2i )2 – 6(3 – 2i ) + 13 ≟ 0
9 – 12i + 4i 2 – 18 + 12i + 13 ≟ 0
4 + 4(–1) ≟ 0
0 = 0
Substitute 3 – 2i for x.
Simplify.
The left side equals the right
side, so 3 – 2i checks.
SQUARE

Solve x2 = 2x + 6 by completing the square.
Solution:
x2 = 2x + 6
x2 – 2x = 6
x2 – 2x + 1 = 6 + 1
(x – 1)2 = 7
Isolate the constant term.
Complete the square.
Factor and simplify.
SOLVE BY COMPLETING THE SQUARE

SOLUTION
x – 1 =
x = 1
The exact solutions of x2 = 2x + 6 are and .
A calculator can be used to show that
and
The decimals –1.646 and 3.646 are approximate solutions of
x2 = 2x + 6.
cont’d
Apply the square root procedure.
Solve for x.

Completing the square by adding the square of half the
coefficient of the x term requires that the coefficient of the
x2 term be 1.
If the coefficient of the x2 term is not 1, then first multiply each
term on each side of the equation by the reciprocal of the
coefficient of x2 to produce a coefficient of 1 for the x2 term.
SOLVE BY COMPLETING THE SQUARE

78.5
8103.4
0372.3
1567.48.2
06426.1144.1
2
2
2
22
22





xx
xx
xx
xxxx
xxxx

SOLVING QUADRATIC EQUATIONS BY USING THE QUADRATIC
FORMULA

Completing the square for ax2 + bx + c = 0 (a  0) produces a
formula for x in terms of the coefficients a, b, and c.
The formula is known as the quadratic formula, and it can
be used to solve any quadratic equation.
The Quadratic Formula
If ax2 + bx + c = 0, a  0, then
As a general rule, you should first try to solve quadratic
equations by factoring. If the factoring process proves difficult,
then solve by using the quadratic formula.
SOLVING QUADRATIC EQUATIONS BY USING THE QUADRATIC
FORMULA

SOLVE BY USING THE QUADRATIC FORMULA
Use the quadratic formula to solve each of the following.
a. x2 = 3x + 5 b. 4x2 – 4x + 3 = 0
Solution:
a. x2 = 3x + 5
x2 – 3x – 5 = 0
Write the equation
in standard form.
Use the quadratic
formula.
a = 1, b = –3, c = – 5.

SOLUTION
b. 4x2 – 4x + 3 = 0
cont’d
The equation is in standard
form.
Use the quadratic formula.

cont’d
a = 4, b = –4, c = 3.
SOLUTION

cont’dSOLUTION

583.3
0181895.584.2
02104.0253.1
2
22
22



xx
xxxx
xxxx

THE DISCRIMINANT OF A QUADRATIC EQUATION

The solutions of ax2 + bx + c = 0, (a  0), are given by
The expression under the radical, b2 – 4ac, is called the
discriminant of the equation ax2 + bx + c = 0.
If b2 – 4ac  0, then is a real number.
If b2 – 4ac  0, then is not a real number.
Thus the sign of the discriminant can be used to determine
whether the solutions of a quadratic equation are real numbers.

The Discriminant and the Solutions of a Quadratic Equation
The equation ax2 + bx + c = 0, with real coefficients and
a  0, has as its discriminant b2 – 4ac.
If b2 – 4ac  0, then ax2 + bx + c = 0 has two distinct real
solutions.
If b2 – 4ac = 0, then ax2 + bx + c = 0 has one real
solution. The solution is a double solution.
If b2 – 4ac  0, then ax2 + bx + c = 0 has two distinct
nonreal complex solutions. The solutions are conjugates
of each other.

USE THE DISCRIMINANT TO DETERMINE THE NUMBER OF REAL
SOLUTIONS
For each equation, determine the discriminant and state
the number of real solutions.
a. 2x2 – 5x + 1 = 0
b. 3x2 + 6x + 7 = 0
c. x2 + 6x + 9 = 0

SOLUTION
a. The discriminant of 2x2 – 5x + 1 = 0 is
b2 – 4ac = (–5)2 – 4(2)(1)
= 17.
Because the discriminant is positive,
2x2 – 5x + 1 = 0 has two distinct real solutions.
b. The discriminant of 3x2 + 6x + 7 = 0 is
b2 – 4ac = 62 – 4(3)(7)
= –48.
Because the discriminant is negative,
3x2 + 6x + 7 = 0 has no real solutions.

c. The discriminant of x2 + 6x + 9 = 0 is
b2 – 4ac = 62 – 4(1)(9)
= 0.
Because the discriminant is 0,
x2 + 6x + 9 = 0 has one real solution.
cont’d
SOLUTION

036.5
04129.4
09124.3
052.2
054.1
2
2
2
2
2





xx
xx
xx
xx
xx
EXERCISES: DETERMINE THE NATURE OF ROOTS OF EACH QUADRATIC
EQUATIONS.

EQUATIONS IN QUADRATIC FORM (OTHER TYPES)
In solving equations in quadratic form, use the following steps:
1.Rewrite the equation in quadratic form using a substitute
variable.
2. Solve the resulting quadratic equation for the substitute
variable.
3. Replace the substitute variable with the original variable and
solve.
4. Check your answer in the original equation.

SOLVE EACH QUADRATIC EQUATION.
1.4
064.3
081.2
9.1
.
3
4
26
35




x
xx
xx
xx
A
0)9(4)9(.4
03649.3
01025.2
01243.1
.
224
246
23
23




xxx
xxx
xxx
xxx
B

4
5
2
1
1
1
.2
5
4
3
2
.32
2
53
.1
.










xx
x
x
x
x
xx
C

22.4
55.3
0
441
.6215.2
065.5212.1
.
23




xx
xx
xxx
xx
xxxx
D

   
089x.4
088x.3
0292-x.2
08
1
16
1
1.1
.
36
24
35
2
















x
x
x
xx
E

     
0x16x8x.4
01x1x51x4.3
06x5x.2
02xx.1
.F
2
1
2
1
2
3
2
5
2
3
2
1
3
2
3
4
6
1
3
1






SUM AND PRODUCT OF ROOTS
Recall from the quadratic formula that when
a2
ac4bb
x0cbxax
2
2 













a2
ac4bb
s
a2
ac4bb
r
sandrberootstheLet
2
2

SUM OF ROOTS
a
b
a2
b2
sr
a2
ac4bbac4bb
a2
ac4bb
a2
ac4bb
22
22









sr
sr
Sum of roots = r + s

PRODUCT OF ROOTS
 
a
c
s)((r)
a4
ac4bb
)s)(r(
a4
ac4b)b(
a2
ac4bb
*
a2
ac4bb
)s(r)(
2
22
2
2
22
22







Product of roots = (r) (s)

EXAMPLE
DETERMINE THE VALUE OF K THAT SATISFIES THE GIVEN CONDITION
 
   
signsopposite
withbutequalynumericallrootsthe0;6-x5-kx1-2k4.
0.isrootstheofone0;65kk10x-x12k3.
1.-isrootsofproduct0;1-k2x2)x(3k2.
20.isrootsofsum0;45xxk1.
2
22
2
2





FINDING THE QUADRATIC EQUATION GIVEN THE ROOTS
   0sxr-x
isequationquadraticthes,andrberootstheLet

i21andi213
27and252
2
1
and
4
3
1


.
.
.
Example: Find the quadratic equations with the given roots.

APPLICATIONS OF QUADRATIC EQUATIONS

A right triangle contains one 90 angle. The side opposite the
90 angle is called the hypotenuse. The other two sides are
called legs.
The lengths of the sides of a right triangle are related by a
theorem known as the Pythagorean Theorem.
The Pythagorean Theorem states that the square of the length of
the hypotenuse of a right triangle is equal to the sum of the
squares of the lengths of the legs.
This theorem is often used to solve applications that involve right
triangles.

The Pythagorean Theorem
If a and b denote the lengths of the legs of a right triangle and c
the length of the hypotenuse, then c2 = a2 + b2 .

A television screen measures 60 inches diagonally, and its aspect
ratio is 16 to 9. This means that the ratio of the width of the
screen to the height of the screen is 16 to 9.
Find the width and height of the screen.
A 60-inch television screen with a 16:9 aspect ratio.
EXAMPLE

SOLUTION
Let 16x represent the width of the screen and let 9x represent
the height of the screen.
Applying the Pythagorean Theorem gives us
(16x)2 + (9x)2 = 602
256x2 + 81x2 = 3600
337x2 = 3600
Solve for x.

 3.268 inches
The height of the screen is about 9(3.268)  29.4 inches, and the
width of the screen is about 16(3.268)  52.3 inches.
cont’d
Apply the square root procedure. The
plus-or-minus sign is not used in this
application because we know x is
positive.
SOLUTION

Quadratic equations are often used to determine the height
(position) of an object that has been dropped or projected.
For instance, the position equation s = –16t 2 + v0t + s0 can be
used to estimate the height of a projected object near the
surface of Earth at a given time t in seconds.
In this equation, v0 is the initial velocity of the object in feet per
second and s0 is the initial height of the object in feet.

SOLVE THE FOLLOWING PROBLEMS.
1. The product of two consecutive positive integers is 56. Find the
lowest integer. ans. 7
2. Paul can finish a job in 2 hrs less than John. Working together they
can finish it in 2 hours and 24 minutes. How long would it take Paul to
finish the job working alone? ans. 4 hours
3.The length of a rectangle is five more than twice its width. The area of
the rectangle is 52 sqm. Find the dimensions of the rectangle.
ans. 4 and 13
4.A tank can be filled in 2 hours by 2 pipes together. The larger pipe can
fill it in 3 hours less than the smaller. How long will it take each pipe to
fill the tank? ans. 6 and 3

5. A car traveled the first 100 kilometers of a trip at one speed and the
last 135 kilometers at an average of 5 kilometers per hour less. If the
entire trip took 5 hours, what was the average speed of the car for
the first part of the trip? ans. 50 kph
6. Find a number whose square exceeds 14 times the number by 51.
7. The denominator of a fraction is 1 more than the
numerator. When 5/6 is added to the fraction, the terms
of the fraction are reversed. What is the original fraction?
8. A boy can row 16 kilometers downstream and return a
distance of 16 kilometers upstream in 5 hours. If the rate
of the current is 6 km/hr, what is the boy’s rate of rowing in
still water?

9. A box is to be formed from a rectangular piece of wood by
cutting equal squares of 10 inches out of the corners and
folding the sides. The piece of wood is thrice as long as it
is wide. If the volume of the box is 7680 in3, What are the
dimensions of the original piece of wood?
10. The science club bought a Php 2400 worth of apparatus for their
experiment. If there had been 8 more students in the club, then
the cost per member to buy the apparatus would have been Php
10 less. How many students are in the club?

MIT Math Syllabus 10-3 Lesson 7: Quadratic equations

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MIT Math Syllabus 10-3 Lesson 7: Quadratic equations