Chapter Closure
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Exercise 138 Page 314
a The expression on the right-side of the equation is given in the standard form of a quadratic function. Start by identifying the values of a, b, and c.
b The expression on the right-side of the equation is given in the standard form of a quadratic function. Start by identifying the values of a, b, and c.
c Use the formula for the difference of two squares.
d Is the given polynomial a perfect square trinomial?
a x=3 or x=4
b x=4/3 or x=5/2
c x=-3 or x=3
d x=- 6
a We want to solve the given equation for x. To do this we can use factoring. Then we will use the Zero Product Property.
Factoring
Here we have a quadratic equation of the form 0=ax^2+bx+c, where |a| = 1 and there are no common factors. To factor this equation we will rewrite the middle term, bx, as two terms. The coefficients of these two terms will be factors of ac whose sum must be b. 0=x^2-7x+12 ⇕ 0= 1x^2+(-7)x+12We have that a= 1, b=-7, and c=12. There are now three steps we need to follow in order to rewrite the above equation.
- Find a c. Since we have that a= 1 and c=12, the value of a c is 1* (12)=12.
- Find factors of a c. Since ac=12, which is positive, factors of a c need to have the same signs — both positive and both negative. Since b=-7, which is negative, both factors must be negative.
c|c|c|c 1^(st)Factor &2^(nd)Factor &Sum &Result - 1 &-12 &- 1 + (-12) &-13 - 2 &-6 &-2 + (-6) &-8 - 3 & -4 & -3 + ( -4) &-7
- Rewrite bx as two terms. Now that we know which factors are the ones to be used, we can rewrite bx as two terms. 0=x^2-7x+12 ⇕ 0=x^2 - 3x - 4x+12
0=x^2-3x-4x+12
FactorOut
Factor out x
0=x(x-3)-4x+12
FactorOut
Factor out -4
0=x(x-3)-4(x-3)
FactorOut
Factor out (x-3)
0=(x-3)(x-4)
Zero Product Property
Since the equation is already written in factored form, we can now use the Zero Product Property.0=(x-3)(x-4)
ZeroProdProp
Use the Zero Product Property
lcx-3=0 & (I) x-4=0 & (II)
AddEqn
(I): LHS+3=RHS+3
lx=3 x-4=0
AddEqn
(II): LHS+4=RHS+4
lx_1=3 x_2=4
b We want to solve the given equation for x. To do this we can use factoring. Then, we will use the Zero Product Property.
Factoring
Here we have a quadratic equation of the form 0=ax^2+bx+c, where |a| ≠ 1 and there are no common factors. To factor this equation we will rewrite the middle term, bx, as two terms. The coefficients of these two terms will be factors of ac whose sum must be b. 0=x^2-7x+12 ⇕ 0= 6x^2+(-23)x+20We have that a= 6, b=-23, and c=20. There are now three steps we need to follow in order to rewrite the above equation.
- Find a c. Since we have that a= 6 and c=20, the value of a c is 6* (20)=120.
- Find factors of a c. Since ac=120, which is positive, factors of a c need to have the same signs — both positive and both negative. Since b=-23, which is negative, both factors must be negative.
c|c|c|c 1^(st)Factor &2^(nd)Factor &Sum &Result - 1 &-120 &- 1 + (-120) &-121 - 2 &-60 &-2 + (-60) &-62 - 3 &-40 &-3 + (-40) &-43 - 4 &-30 &-4 + (-30) &-34 - 5 &-24 &-5 + (-24) &-29 - 6 &-20 &-6 + (-20) &-26 - 8 & -15 & -8 + ( -15) &-23 - 10 &-12 &-10 + (-12) &-22
- Rewrite bx as two terms. Now that we know which factors are the ones to be used, we can rewrite bx as two terms. 0=6x^2-23x+20 ⇕ 6x^2 - 8x - 15x+20
0=6x^2-8x-15x+20
FactorOut
Factor out 2x
0=2x(3x-4)-15x+20
FactorOut
Factor out -5
0=2x(3x-4)-5(3x-4)
FactorOut
Factor out (3x-4)
0=(3x-4)(2x-5)
Zero Product Property
Since the equation is already written in factored form, we can now use the Zero Product Property.0=(3x-4)(2x-5)
ZeroProdProp
Use the Zero Product Property
lc3x-4=0 & (I) 2x-5=0 & (II)
lcx= 43 & (I) 2x-5=0 & (II)
lx_1= 43 x_2= 52
Applying the Difference of Two Squares
Do you notice that the expression on the left-hand side of the equation is a difference of two perfect squares? This can be factored using the difference of squares method.a^2 - b^2 ⇔ (a+b)(a-b) To do so, we first need to express each term as a perfect square.
| Expression | x^2-9 |
|---|---|
| Rewrite as Perfect Squares | (x)^2 - 3^2 |
| Apply the Formula | (x+3)(x-3) |
Solving the Equation
Finally, to solve the equation we will use the Zero Product Property.x^2-9=0
▼
a^2-b^2=(a+b)(a-b)
(x+3)(x-3)=0
ZeroProdProp
Use the Zero Product Property
lcx+3=0 & (I) x-3=0 & (II)
lx=- 3 x=3
d We want to solve the given equation for x. To do this we can start by using factoring. Then, we will use the Zero Product Property.
Factoring
We want to factor the given equation. Factoring is much easier when our polynomial is a perfect square trinomial. To determine if an expression is a perfect square trinomial, we need to ask ourselves three questions.| Is the first term a perfect square? | x^2= x^2 ✓ |
| Is the last term a perfect square? | 36= 6^2 ✓ |
| Is the middle term twice the product of 6 and x? | 12x=2* 6* x ✓ |
As we can see, the answer to all three questions is yes! Therefore, we can write the trinomial as the square of a binomial. Note there is an addition sign in the middle. x^2+12x+36=0 ⇔ ( x+ 6)^2=0
Zero Product Property
Since the equation is already written in factored form, we can now use the Zero Product Property.(x+6)^2=0
PowToProdTwoFac
a^2=a* a
(x+6)(x+6)=0
ZeroProdProp
Use the Zero Product Property
lcx+6=0 & (I) x+6=0 & (II)
(I), (II): LHS+6=RHS+6
lx_1=-6 x_2=-6
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