# Math Help - Solving for x & side length in a right triangle

1. ## Solving for x & side length in a right triangle

Hello,
I have to solve for x and side lengths in a right triangle.
1. The two equidistant sides are x and x + 6
2. The hypotenuse is 2 sqrt 17

If I am correct so far, I used the Pythagorean Theorem. So:

x^2 + (x + 6)^2 = (2 sqrt 17)^2

x^2 + x^2 + 36 = I don't know how to do this.

Thank you very much!

2. Important thing to note. In general: $(a+b)^{2}\: {\color{red}\neq} \: a^{2} + b^{2}$

You have to multiply it out with the FOIL method you've learned at school: $(a+b)^{2} = (a+b)(a+b) = ... = a^{2} + 2ab + b^{2}$

Recall that: $(ab)^{c} = a^{c}b^{c}$

So for your radical term, imagine a = 2 and b = sqrt{17}.

3. Hello,

Originally Posted by VAP
Hello,
I have to solve for x and side lengths in a right triangle.
1. The two equidistant sides are x and x + 6
2. The hypotenuse is 2 sqrt 17

If I am correct so far, I used the Pythagorean Theorem. So:

x^2 + (x + 6)^2 = (2 sqrt 17)^2

x^2 + x^2 + 36 = I don't know how to do this.

Thank you very much!
The problem is :

$(x+6)^2=x^2+2 \cdot 6 \cdot x+36$

$(2 \sqrt{17})^2=2^2 \cdot (\sqrt{17})^2=4 \cdot 17=68$

Therefore :

$x^2+x^2+12x+36=68$

$2x^2+12x-32=0$

Divide by 2 :

$x^2+6x-16=0$

Complete the square :

$x^2+6x+3^2-3^2-16=0$

$(x+3)^2-25=(x+3)^2-5^2=0$

Using the identity : $a^2-b^2=(a-b)(a+b)$ :

$(x+3-5)(x+3+5)=0$

Continue... or finish

4. Originally Posted by Moo
$(x+6)^2=x^2+2 \cdot 6 \cdot x+36$

$(2 \sqrt{17})^2=2^2 \cdot (\sqrt{17})^2=4 \cdot 17=68$

Therefore :

$x^2+x^2+12x+36=68$

$2x^2+12x-32=0$

Divide by 2 :

$x^2+6x-16=0$

Complete the square :

$x^2+6x+3^2-3^2-16=0$

$(x+3)^2-25=(x+3)^2-5^2=0$

Using the identity : $a^2-b^2=(a-b)(a+b)$ :

$(x+3-5)(x+3+5)=0$

Continue... or finish
-----------------------------------
Moo, I liked your approach to this.

I would just say that when you get to $x^2+6x-16=0$,
try to factor the trinomial either "guess and check" or this way:

Ask yourself, 'what 2 factors when multiplied together give you -16 and when added give you +6 (the coef. of the middle term)?'

Not many factors of 16 to worry with, so let's try $+8$ and $-2$. I believe that'll do the trick.

$(x+8)(x-2)=0$

The zero product property tell us that $x+8=0$, or $x-2 - 0$

Then, $x=-8$, or $x=2$

$x=-8$ is extraneoous since the lengths cannot be negative.

Therefore, $x=2$ and $x+6=8$

5. Hi masters

Actually, I use this method for explaining to people... I would use what you've described, or just calculating the discriminant quickly to see what the solutions look like, if I had to deal with the problem

Thanks for noticing !

(VAP : please send mails instead of PMs, it will be a pleasure )

--- This is my 2000th post

6. Originally Posted by Moo
Hi masters

--- This is my 2000th post

CONGRATULATIONS on a great milestone....one that I may never reach. Keep up the great postings.

Dale