Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
4. Factoring to Solve Quadratic Equations
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Exercise 42 Page 572

Practice makes perfect
a We can write a quadratic equation in factored form using the given roots. Then we will change it to standard form by multiplying the factors.
Factored Form:& a(x-p)(x-q)=0 Standard Form:& ax^2+bx+c=0 In the factored form, p and q are the roots of the equation. Since we are told the roots are - 5 and 8, we can partially write the factored form of our equation. a(x-( - 5))(x-8)=0 ⇕ a(x+5)(x-8)=0 Since a does not have any effect on the roots, we can choose any value. For simplicity and in order to have integer coefficients, we will let a=1. 1(x+5)(x-8)=0 ⇕ (x+5)(x-8)=0 Finally, let's use the Distributive Property to obtain the standard form.
(x+5)(x-8)=0
â–Ľ
Multiply parentheses
x(x-8)+5(x-8)=0
x^2-8x+5(x-8)=0
x^2-8x+5x-40=0
x^2-3x-40=0
Please note that this is just one example of a quadratic function that satisfies the given requirements.
b We can write a quadratic equation in factored form using the given roots. Then we will change it to standard form by multiplying the factors.
Factored Form:& a(x-p)(x-q)=0 Standard Form:& ax^2+bx+c=0 In the factored form, p and q are the roots of the equation. Since we are told the roots are 3 and - 2, we can partially write the factored form of our equation. a(x-3)(x-(- 2))=0 ⇕ a(x-3)(x+2)=0 Since a does not have any effect on the roots, we can choose any value. For simplicity and in order to have integer coefficients, we will let a=1. 1(x-3)(x+2)=0 ⇕ (x-3)(x+2)=0 Finally, let's use the Distributive Property to obtain the standard form.
(x-3)(x+2)=0
â–Ľ
Multiply parentheses
x(x+2)-3(x+2)=0
x^2+2x-3(x+2)=0
x^2+2x-3x-6=0
x^2-x-6=0
Please note that this is just one example of a quadratic function that satisfies the given requirements.
c We can write a quadratic equation in factored form using the given roots. Then we will change it to standard form by multiplying the factors.
Factored Form:& a(x-p)(x-q)=0 Standard Form:& ax^2+bx+c=0 In the factored form, p and q are the roots of the equation. Since we are told the roots are 12 and - 10, we can partially write the factored form of our equation. a(x-1/2)(x-(- 10))=0 ⇕ a(x-1/2)(x+10)=0 Since a does not have any effect on the roots, we can choose any value. For simplicity and in order to have integer coefficients, we will let a=2. This will allow us to eliminate the fractions when we distribute. 2(x-1/2)(x+10)=0 Finally, let's use the Distributive Property to obtain the standard form.
2(x-1/2)(x+10)=0
( 2x-1 ) (x+10)=0
â–Ľ
Multiply parentheses
2x(x+10)-1(x+10)=0
2x^2+20x-1(x+10)=0
2x^2+20x-x-10=0
2x^2+19x-10=0
Please note that this is just one example of a quadratic function that satisfies the given requirements.
d We can write a quadratic equation in factored form using the given roots. Then we will change it to standard form by multiplying the factors.
Factored Form:& a(x-p)(x-q)=0 Standard Form:& ax^2+bx+c=0 In the factored form, p and q are the roots of the equation. Since we are told the roots are 23 and - 57, we can partially write the factored form of our equation. a( x-2/3 )( x-( - 5/7 ) )=0 ⇕ a( x- 2/3 ) ( x+5/7 )=0 Since a does not have any effect on the roots, we can choose any value. For simplicity and in order to have integer coefficients, we will let a=21. This is a common multiple of both denominators of the given roots and will allow us to eliminate the fractions when we distribute. 21( x- 2/3 ) ( x+5/7 )=0 Finally, let's use the Distributive Property to obtain the standard form.
21( x- 2/3 ) ( x+5/7 )=0
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Distribute 21
3(7)( x- 2/3 ) ( x+5/7 )=0
3( x-2/3 ) (7)( x+5/7 )=0
( 3x- 2 ) (7)( x+5/7 )=0
( 3x-2 ) (7x+5)=0
â–Ľ
Multiply parentheses
3x(7x+5)-2(7x+5)=0
21x^2+15x-2(7x+5)=0
21x^2+15x-14x-10=0
21x^2+x-10=0
Please note that this is just one example of a quadratic function that satisfies the given requirements.