By the way, how was I right?

Question

My math test had this question:

Solve for $x$: $$\sqrt{8x-12}+\sqrt{32x+52}=10.$$

I solved it as follows: $$\sqrt{8x-12}+\sqrt{32x+52}=10.$$ Dividing both sides by 2, $$\sqrt{2x-3}+\sqrt{8x+13}=5.$$ squaring on both sides, $$(\sqrt{2x-3}+\sqrt{8x+13})^2=25$$ Cancelling square and square roots, we get $$(2x-3+8x+13)=25$$ $$10x=15$$ $$x=1.5$$

The answer key had the same anwer. So I expected an 'A' grade .

But the teacher gave me 'F' and wrote "Right answer but wrong method" in my answer sheet.

Can anyone explain why I got the right answer inspite of doing it in a wrong way?

Answer 1

The mistake is at the step: $$\left (\sqrt{2x-3}+\sqrt{8x+13}\right )^2=25$$ Which you, basicly wrote, is equal to: $$(\sqrt{2x-3})^2+(\sqrt{8x+13})^2=25$$

The reason why this gave the right answer, is because $x=1.5$, and so $\sqrt{2x-3}=0$.

In other words, $(a+b)^2=a^2+b^2$ if $a=0$.

Answer 2

Note that $$(a+b)^2 = a^2 +b^2 +2ab$$

You got it right despite doing $a^2 + b^2$ because in your case $2ab = 0$ i.e. $2(1.5) - 3 =0$.