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