Solving a word problem using a quadratic equation with rational root

Question

The length of a rectangle is 7 ft² more than double the width and the area of the rectangle is 99 ft² find the dimensions rectangle

Answer 1

First of all, I think you mean that the length is 7 feet more than twice the width, not 7 square feet. Length is a one dimensional measurement. Square feet (ft²) is a two-dimenstional measurement, as it involves length AND width.

Given this assumption, let's say that length= L=2w+7. Area, of course, is two dimensions, length times width: A=L· w. Substitute the formula for length with w in it into the area formula: A=(2w+7)w and distribute the w to get: A=2w²+7w. The question states that A=99, so plug that in: 99=2w²+7w. Subtract 99 from both sides to get the equation in standard form: 0=2w²+7w-99. Use the quadratic formula to find the answers. (If you don't know the quadratic formula, look it up and MEMORIZE IT!)

Given 2w²+7w-99, we have values a=2, b=7, and c= -99.

Plugging them into the quadratic formula, I get w= -9, 5.5.

We can't have a negative width, so w=5.5.

Plugging that into the formula for length, we get L=2(5.5)+7=11+7=18. So L=18.

I hope this helps, and please let me know if you have any more questions.

Have fun mathing!

Liz Z.

Answer 2

l = 7 + 2w

A = 99 = wl = w(7 + 2w)

Now, solve the quadratic 2w² + 7w - 99 = 0

Should be 7 ft, not 7 ft²

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