
Quadratics
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.
The answer is (15+30 root2)/7
But I got this 5+root50 through the following working:
Could you please tell me where I went wrong?
http://i83.photobucket.com/albums/j283/ta_lia/ahhh.jpg

The right hand side should be $\displaystyle 9x^2$


Because both the 3 and x have to be squared:
$\displaystyle \begin{aligned}
x^2 + (x + 5)^2 &= (3x)^2 \\
x^2 + x^2 + 10x + 25 &= 9x^2 \\
2x^2 + 10x + 25 &= 9x^2 \\
0 &= 7x^2  10x  25
\end{aligned}$
Then solve using the quadratic formula.
01

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
Use quad formula from here to get answer.

Oh could you tell me how I messed up in the quadratic formula then?
http://i83.photobucket.com/albums/j2...lia/img053.jpg
^ My answer is (5+/ 10root2)/7 rather than ( 5+/ 30root2 )/7

7x²10x25=0
Using quad
= (10) + root of (10)²  4 (7) (25)
Divided by 2(7)
=10 + root of 100 + 700
divided by 14

Firstly:
why is it 100700 > 4 * 7 *25 should be positive since there are two negatives multiplying against one another
Secondly:
if it is 100700 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


$\displaystyle 7x^2  10x  25 = 0$
$\displaystyle \begin{aligned}
x &= \frac{10 \pm \sqrt{(10)^2  4(7)(25)}}{2(7)} \\
&= \frac{10 \pm \sqrt{100 + 700}}{14} \\
&= \frac{10 \pm \sqrt{800}}{14} \\
&= \frac{10 \pm 20\sqrt{2}}{14} \\
&= \frac{5 \pm 10\sqrt{2}}{7}
\end{aligned}$
Hmm, I'm not getting the original answer. I'm getting what prettiestfriend got.
01