Simba9 wrote:

Hi

BunuelSince you say it's better to solve (when we get an equation like a^2 − 32a −32=0, as we cannot assume that one value will be positive and the other neg.) Would you mind posting a complete resolution to this problem? i.e. including your calculations to get to the positive and negative (invalid) solution.

I am asking this as I think you might be doing the calculus in a much more straightforward way than I am.

Thanks in advance!

Here you go!! I know it's a bit late but it might be useful for others as well.

\(a^2 -32a-32=0\). You can't solve this using factorization. So we should use other method

\(D = \sqrt{b^2-4*a*c}\)

\(D = \sqrt{(-32)^2-4*1*(-32)}\)

\(D = \sqrt{(32 * 32) + (4*32)}\)

\(D = \sqrt{32 * (32 + 4)}\)

\(D = \sqrt{32 * 36}\)

\(D = 24 \sqrt{2}\)

\(D = 34\) approx.

\(x = \frac{(-b + D)}{(2*a)}\)

\(a =\frac{(- - 32 + 34)}{(2*1)}\)

\(a = \frac{(32 + 34)}{2}\)

\(a = 33\)

\(x = \frac{(-b - D)}{(2* a)}\)

\(a = \frac{(- - 32 - 34)}{(2*1)}\)

\(a = \frac{(32 - 34)}{2}\)

\(a = \frac{-2}{2}\) = -1 Not Applicable.

so a = 33, we can calculate value of b. And Perimeter of Rectangle as well.

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Thank You

Sudhanshu