Exercise 10 Page 271

Let's start with reviewing the formula for the area of a rectangle. A=wl Here, w is the width of the rectangle and l is the length of the rectangle. We are told that an area of the given rectangle is 48 square units. Let's substitute this value for A into the formula. 48=wl From the diagram, we know that the rectangle's width w is x-8 and the length l is x. 48=( x-8) x Now, we can solve the equation we got for x.

48=(x-8)x

Distr

Distribute x

48=x^2-8x

SubEqn

LHS-48=RHS-48

0=x^2-8x-48

RearrangeEqn

Rearrange equation

x^2-8x-48=0

This equation is quadratic. Let's try to solve it using factoring!

x^2-8x-48=0

WriteSum

Write as a sum

x^2-(12x-4x)-48=0

Distr

Distribute -1

x^2-12x+4x-48=0

▼

Factor out x & 4

FactorOut

Factor out x

x(x-12)+4x-48=0

FactorOut

Factor out 4

x(x-12)+4(x-12)=0

FactorOut

Factor out x-12

(x+4)(x-12)=0

ZeroProdProp

Use the Zero Product Property

lx_1=- 4 x_2=12

We got two possible values of x. The first one is negative. However, let's recall that on the diagram x is the measure of the rectangle's length, which cannot be negative. Therefore, the value of x is 12.