I am trying to solve below question from Coursera Intro to Calculus (link)

A right-angled triangle has shorter side lengths exactly $a^2-b^2$ and $2ab$ units respectively, where $a$ and $b$ are positive real numbers such that $a$ is greater than $b$. Find an exact expression for the length of the hypotenuse (in appropriate units).

Below are the choices

$(a - b)^2$

$\sqrt(a^4 + 4a^2b^2 -b^4)$

$a^2 + b^2$

$\sqrt(a^2 + 2ab -b^2)$

$(a + b)^2$

When I attempt to work out the solution (and I even got a 2nd pair of eyes to look at it, but he arrived at the same conclusion), I get this:

$(a^2 - b^2)^2 + (2ab)^2 = x^2$

$a^4 -2a^2b^2 + b^4 + (2ab)^2 = x^2$

$a^4 -2a^2b^2 + b^4 + 4a^2b^2 = x^2$

$a^4 + 2a^2b^2 + b^4 = x^2$

$\sqrt(a^4 + 2a^2b^2 + b^4) = x$

Please help! How to get the correct solution?


You are done. just multiply out $(a^2 + b^2)^2$ and check that you get $x^2$.


$$a^4+2a^2b^2+b^4 = (a^2+b^2)^2$$


Let $$a' = a^{2} - b^{2}$$ $$ b' = 2ab $$

From the Pythagorean Theorem, $$ c'^{2} = a'^{2} + b'^{2} $$

Substitute both $a'$ and $b'$ into the above equation:

$$ c'^{2} = (a^{2} - b^{2})^{2} + (2ab)^{2} = a^{4} - 2a^{2}b^{2} + b^{4} + 4a^{2}b^{2} = a^{4} + 2a^{2}b^{2} + b^{4}$$

Solve for $c'$ by taking the square root:

$$ c' = \sqrt{a^{4} + 2a^{2}b^{2} + b^{4}} $$

Here, $a^{4} + 2a^{2}b^{2} + b^{4}$ is a perfect square, and $$ \sqrt{a^{4} + 2a^{2}b^{2} + b^{4}} = \sqrt{(a^2 + b^2)^{2}} = a^2 + b^2$$

  • $\begingroup$ Is the minus sign on the last term of Option 2 not relevant? $\endgroup$ – DJohnM May 30 at 16:49
  • $\begingroup$ Thank you for the correction. I will edit my answer accordingly. $\endgroup$ – Marvin May 30 at 16:52

$$c^2=(a^2-b^2)^2+(2ab)^2$$ $$=(a^4-2a^2b^2+b^4)+(4a^2b^2)$$ $$=a^4+2a^2b^2+b^4=(a^2+b^2)^2$$ $$\Rightarrow c=a^2+b^2$$


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