Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
2. Section 5.2
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Exercise 101 Page 296

Practice makes perfect
a Stanton is talking about the Zero Product Property. When x equals 5 or -2, the left-hand side will equal 0. However, a solution to an equation has to make the left-hand side and right-hand side equal. This is not the case for the given equation and proposed solutions.

( 5-5)( 5+2)? =-6 & ⇔ 0 ≠ -6 * ( -2-5)( -2+2)? =-6 & ⇔ 0 ≠ -6 * As we can see, the proposed solutions does not result in true statements. Therefore, these values of x cannot be solutions to the equation.

b To rewrite the equation, we must make sure that the right-hand side equals 0.
(x-5)(x+2)=-6
x^2+2x-5x-10=-6
x^2-3x-10=-6
x^2-3x-4=0
Now the right-hand side is 0. To factorize the left-hand side, we can use a generic rectangle and a diamond problem. We know that x^2 and -4 goes into the lower left and upper right corners of the generic rectangle.

To fill in the remaining two corners, we need two x-terms that have a sum of -3x and a product of - 4x^2.

Notice that the product is negative. This means one factor must be positive and the other must be negative. |c|c|c|r|c| [-1em] Product & ax(bx) & ax+bx & Sum & -3x? [0.2em] [-1em] -4x^2 & - x(4x) &- x+4x& 3x & * [0.1em] -4x^2 & - 4x(x) & -4x+x& -3x & âś“ [0.1em] When one term is - 4x and the other is x, we have a product of -4x^2 and a sum of -3x. Now we can complete the diamond and generic rectangle.

To factor the left-hand side we add each side of the generic rectangle and multiply the sums. x^2-3x-4=0 ⇓ (x+1)(x-4)=0

c Since the right-hand side now equals zero, we can solve the equation with the Zero Product Property.
(x+1)(x-4)=0
lcx+1=0 & (I) x-4=0 & (II)
lx=-1 x-4=0
lx_1=-1 x_2=4
The solutions are x=-1 and x=4, which do not match with Kyle's solutions.