The length of a rectangle is 4cm more than the width the area is 96cm

^{2 }-
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The length of a rectangle is 4cm more than the width the area is 96cm^{2 }

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In this problem we have the length (l) and the area (A) of a triangle. The length is 4 more than the width. we can have the width be represented by w. Therefore, if the length of the triangle is 4 more than the width, the length would be:
**w + 4**.

The equation for finding the area of a rectangle is: **A = lw**

If if I fill out the equation with our filler values, we get: 96cm^{2} = (w+4)(w). Then you solve -

96cm^{2} = (w+4)(w)

96 = w^{2 }+ 4w

0 = w^{2} + 4w - 96

Oh hey, quadratic equation! First we need to separate it into two parts.

0 = (w + )(w - ) Cool. Now we need two numbers whose product = 96, and whose difference = 4.

0 = (w + 12)(w - 8) Awesome. Now set each side equal to 0 and solve. Why? Because anything times 0 equals 0. Therefore, the equation will be true as long as one of those two parts is equal to 0.

w + 12 = 0

w = -12

w - 8 = 0

w = 8

w = -12 ; w = 8. Only one of these answers is possible for a measurements. You can't have negative length measurements because then you're traveling into other dimensions. And if you are, please take picture, that is groundbreaking science.

So there's the width! w = 8. So what's the length again? Oh right 4 more than the width.

l = 12 and w = 8.

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