Section 15-1: Solving Quadratic Equations by the Square-Root Method
Learning Outcome 1
Write the equation 7 = 13x + 8x2 in standard form.
7 = 13x + 8x2
Rearrange the equation so that all terms are on the left and
arranged in descending powers.
2
−8x − 13x + 7 = 0
Factor negative one from the three terms on the left.
2
−1(8x + 13x − 7) = 0
Multiply both sides of the equation by –1 so that the leading
8x2 + 13x − 7 = 0
coefficient is positive.
Learning Outcome 2
Rewrite the equation 7x − 4x2 = −5 in standard form, then identify the constant term, the
coefficient of the linear term, and the coefficient of the quadratic term.
7x − 4x2 = −5
Rearrange the equation so all terms on the left of the equal sign,
the terms are in descending powers of x, and the leading coefficient
is positive.
2
Next, identify the various coefficients. The constant term is the term
4x − 7x + 5 = 0
that has no letter, 5. The linear term is the term that has the letter x
with no exponent (understood exponent of 1). The linear term is
–7x and its coefficient is –7. The quadratic term is the term that has
the letter “squared.” This term is 4 x 2 and the coefficient is 4.
Learning Outcome 3
Solve: 5x2 − 20 = 0
5x2 − 20 = 0
5x2 = 20
x2 = 4
x = ±2
Solve for x2. Add 20 to both sides of equation.
Divide both sides of equation by 5.
Simplify.
Take the square root of both sides of the equation.
Section 15-2: Solving Quadratic Equations by Factoring
Learning Outcome 1
Solve: 3x2 − 21x = 0
3x2 − 21x = 0
3x(x − 7) = 0
3x = 0; x – 7 = 0
x = 0; x = 7
Be sure the equation equals zero. Find the common factor.
Set each factor equal to zero.
Solve each equation for x.
Learning Outcome 2
Use factoring to solve: 6x2 - 19x + 10 = 0
6x2 - 19x + 10 = 0
Factor trinomial into two binomials.
(2x – 5)(3x –2) = 0
Set each factor equal to zero.
2x – 5 = 0 3x –2 = 0
2x = 5 3x = 2
5
2
x=
x=
2
3
Solve each equation for x.
Divide by coefficient of x.
Section 15-3: Solving Quadratic Equations by Completing the Square
Solve by completing the square.
3 x 2 − 2x − 6
Isolate variable terms.
3 x 2 − 2x = 6
Divide by leading coefficient.
3x 2 2
6
− x=
3
3
3
2
x2 − x = 2
3
62
4
1
1
=
= . Add
to both sides.
2
9 (4) 9
9
4a
2
1
1
x2 - x + = 2 +
3
9
9
Factor left side as a binomial square.
2
1
18 1
x=
+
9 9
Ł 3ł
2
Take square root of both sides.
1
19
=–
3
9
Simplify.
1
19
x=
9
Ł 3ł
x-
Combine fractions.
x=
1
19
–
3
3
Section 15-4: Solving Quadratic Equations Using the Quadratic
Formula
Learning Outcome 1
Use the quadratic equation to solve for x: 4x2 − 6x + 2 = 0
First identify a, b, and c. a = 4; b = –6; c = 2
− b ± b 2 − 4ac
x=
2a
Substitute the values for a, b, and c, respectively.
− ( −6) ± ( −6) 2 − 4( 4)(2)
x=
2( 4)
Perform operations to simplify radicand.
6 ± 36 − 32
8
6± 4
x=
8
6±2
x=
8
6+2
6−2
x=
and x =
8
8
x=
Evaluate
4.
Interpret the ± symbol and find the two roots.
Simplify and reduce.
8
4
and x =
8
8
1
x =1
x=
2
x=
Find the length and width of a rectangular land plot if the length is 40 meters longer than the width
and the area is 48,000 square meters. Let x = number of meters in the width
x + 40 = number of meters in the length
Use the formula for finding the area of a rectangle: A = lw or lw=A
lw = A
Substitute known and unknown values.
(x + 40)(x) = 48,000
Multiply, then write the equation in standard form.
2
x + 40x − 48,000 = 0
Factor if possible or use the quadratic formula.
(x + 240)(x − 200) = 0
Factor, then set each factor equal to zero to solve for x.
x + 240 = 0 or x – 200 = 0
Solve each of the equations.
x = –240; x = 200
Measurements cannot be negative. Therefore, the width is 200, and
the length, x + 40, is 200 + 40 or 240.
Learning Outcome 2
Use the discriminant to determine the characteristics of the roots of the equation: 2x2 + 5x −8 = 0
Identify a, b, and c.
2x2 + 5x −8 = 0
a = 2; b = 5; c = –8
Write the discriminant of the quadratic equation.
b2 − 4ac
Substitute values for variables in discriminant.
2
5 − 4(2)(−8)
Simplify using order of operations.
25 + 64
Add.
89
Because 89 is positive and not equal to zero, there will be two roots
and they will be real. Because 89 is not a perfect square, the roots
will be irrational and unequal.
Section 15-5: Solving Higher-Degree Equations by Factoring
Learning Outcome 1
Determine the degree of the equation: 7x2 − 4x5 + 2 = 0
7x2 − 4x5 + 2 = 0
The degree of the first term is 2, the degree of the second term is 5,
and the degree of the third term is zero. Therefore, the degree of
the equation is the highest of these degrees, 5.
Learning Outcome 2
Solve the equation by factoring: x3 − 7x2 + 10x = 0
x3 − 7x2 + 10x = 0
Factor the common factor x first.
2
x(x − 7x + 10) = 0
Factor the trinomial.
x(x − 5)(x − 2) = 0
Set each variable factor equal to zero.
x = 0; x – 5 = 0; x – 2 = 0
Solve each equation.
x = 0; x = 5; x = 2