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Tutorial 3.7 Solve quadratic equations by factoring
Just as we solved linear equations (standard form: Ax + By = C; slope-intercept form: y = mx + b) to complete some business applications, we may also need to solve quadratic equations (ax2 + bx + c = 0) to complete other such applications.
One way to solve quadratic equations is through factoring and the application of:
Zero Product Property: The product of any two real numbers will equal zero if either or both of these numbers is equal to zero. Expressed differently, if (a)(b) = 0, then a = 0 (alone), b = 0 (alone), or both a and b equal zero.
example 3.7a Solve by factoring: 16x2 – 25 = 0
1. Set the quadratic equation to zero. (This step is already completed.) 2. Factor: 16x2 – 25 → Since this binomial is the difference of squares,
we can use the “special pattern” to complete the factoring:
a2 – b2 = (a + b)(a – b)
16x2 – 25 = (4x)2 – (5)2 = (4x + 5)(4x – 5) 3. Set each factor equal to zero and solve each “mini-equation”:
4x + 5 = 0 → 4x = –5 → x = –5/4 = –1.25 4x – 5 = 0 → 4x = +5 → x = +5/4 = +1.25
4. Determine the solutions for the given quadratic: x = ±5/4 = ± 1.25 example 3.7b Solve by factoring: x2 – 4x = 5x2
1. Set the quadratic equation to zero:
x2 – 4x = 5x2→ 0 = 5x2 – x2 + 4x = 4x2 + 4x
2. Factor: 4x2 + 4x → Factor out the greatest common factor: 4x(x + 1) 3. Set each factor equal to zero and solve each “mini-equation”:
4x = 0 → x = 0/4 = 0 x + 1 = 0 → x = –1
4. Determine the solutions for the given quadratic: x = 0, x = –1
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example 3.7c Solve by factoring: x2 + 12x = 4x – 15 1. Set the quadratic equation to zero:
x2 + 12x = 4x – 15 → x2 + 8x + 15 = 0
2. Factor: x2 + 8x + 15 → We are looking for two numbers that multiply to be +15 and add to be +8: +3, +5 →
x2 + 8x + 15 = x2 + 3x + 5x + 15 = x(x + 3) + 5(x + 3) = (x + 3)(x + 5) 3. Set each factor equal to zero and solve each “mini-equation”:
x + 3 = 0 → x = –3 x + 5 = 0 → x = –5
4. Determine the solutions for the given quadratic: x = –3, x = –5 example 3.7d Solve by factoring: 25x2 + 20x + 4 = 0
1. Set the quadratic equation to zero. (This step is already completed.)
2. Factor: 25x2 + 20x + 4 → We are looking for two numbers that multiply to be +100 (25 × 4 = 100) and add to be +20: +10, +10. Since these numbers are identical, the trinomials is a perfect square and we can factor it by using one of the “special patterns”: a2 + 2ab + b2 = (a + b)2→
25x2 + 20x + 4 = (5x)2 + 2(5)(2)x + (2)2 = (5x + 2)2
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3. Set each factor equal to zero and solve each “mini-equation”:
5x + 2 = 0 → 5x = –2 → x = –2/5 = –0.4
NOTE: Since we have a perfect square trinomial, we will have a “double root” as a solution and only one “mini-equation” needs to be solved.
4. Determine the solution for the given quadratic: x = –2/5 = –0.4 example 3.7e Solve by factoring: 4 2 7 1
15 15 x + x =
1. Set the quadratic equation to zero: 4 2 7 1 15 15
x + x= → 4 2 7 1 0 15 15
x + x− =
2. Because of the fractions, multiply through by the lowest common denominator (which is 15 for this example): 4x2 + 7x – 15 = 0
3. Factor: 4x2 + 7x – 15 → We are looking for two numbers that multiply to be –60 (–15 × 4 = –60) and add to be +7: +12, –5 →
4x2 + 7x – 15 = 4x2 + 12x – 5x – 15 = 4x(x + 3) – 5(x + 3)
= (x + 3)(4x – 5)
4. Set each factor equal to zero and solve each “mini-equation”:
x + 3 = 0 → x = –3 4x – 5 = 0 → 4x = 5 → x =5
4= 1.25
5. Determine the solutions for the given quadratic: x = –3, x = 5/4 = 1.25 example 3.7e Solve by factoring: (x – 1)(2x + 1) = 14
1. Multiply the two binomials: (x – 1)(2x + 1) = 2x2 + x – 2x – 1 = 2x2 – x – 1 2. Set the quadratic equation to zero: 2x2 – x – 1= 14 → 2x2 – x – 15 = 0 3. Factor: 2x2 – x – 15 → We are looking for two numbers that multiply to
be –30
(–15 × 2 = –30) and add to be –1: +5, –6.
2x2 – x – 15 = 2x2 – 6x + 5x – 15 = 2x(x – 3) + 5(x – 3) = (x – 3)(2x + 5) 4. Set each factor equal to zero and solve each “mini-equation”:
x – 3 = 0 → x = 3 2x + 5 = 0 → 2x = –5 → x = 5
2
− = –2.5
5. Determine the solutions for the given quadratic: x = 3, x = –5/2 = –2.5
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