The question:

In a right angled triangle the second to longest side is 5cm longer than the shortest and the hypotenuse is 3 times longer than the shortest side. Find the exaact length of the hypotenuse.

But I got this 5+root50 through the following working:
Could you please tell me where I went wrong?

2. The right hand side should be $9x^2$

3. umm why?

4. Because both the 3 and x have to be squared:

\begin{aligned}
x^2 + (x + 5)^2 &= (3x)^2 \\
x^2 + x^2 + 10x + 25 &= 9x^2 \\
2x^2 + 10x + 25 &= 9x^2 \\
0 &= 7x^2 - 10x - 25
\end{aligned}

Then solve using the quadratic formula.

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5. Your working should be something like this

a² + b² = c²
(x)²+ (5+x)² = (3x)²
x² + 25 + 10x + x² = 9x²
2x² +10x + 25 =9x²
7x²- 10x -25 = 0

6. Oh could you tell me how I messed up in the quadratic formula then?

^ My answer is (5+/- 10root2)/7 rather than ( 5+/- 30root2 )/7

7. 7x²-10x-25=0

= -(-10) + root of (-10)² - 4 (7) (-25)
Divided by 2(7)

=10 + root of 100 + 700
divided by 14

8. Firstly:
why is it 100-700 --> -4 * 7 *-25 should be positive since there are two negatives multiplying against one another

Secondly:
if it is 100-700 then it will be a NEGATIVE number which cannot square root and therefore the answer will be "no real solutions" which is even more incorrect

9. Um how did you get that?

10. $7x^2 - 10x - 25 = 0$

\begin{aligned}
x &= \frac{10 \pm \sqrt{(-10)^2 - 4(7)(-25)}}{2(7)} \\
&= \frac{10 \pm \sqrt{100 + 700}}{14} \\
&= \frac{10 \pm \sqrt{800}}{14} \\
&= \frac{10 \pm 20\sqrt{2}}{14} \\
&= \frac{5 \pm 10\sqrt{2}}{7}
\end{aligned}

Hmm, I'm not getting the original answer. I'm getting what prettiestfriend got.

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