Question

How do I solve for x

Answer 1

Greetings!

To find the length of any side of a right triangle, you can use the Pythagorean Thereom. It states that the squares of two sides are equal to the square of the hypotenuse:
$a^2+b^2=c^2$

Input the information from the diagram into the formula: 
$(x)^2+(x+7)^2=(13)^2$

Expand each term:
$(x)^2+(x+7)^2=(13)^2$

$x^2+((x+7)(x+7))=169$

$x^2+(x(x+7)+7(x+7))=169$

$x^2+x^2+7x+7x+49=169$

Combine like terms:
$2x^2+14x+49=169$

Add -169 to both sides:
$(2x^2+14x+49)+(-169)=(169)+(-169)$

$2x^2+14x-120=0$

Factor out the Common Term (2):
$2(x^2+7x-60)=0$

Factor the Complex Trinomial:
$2(x^2-5x+12x-60)=0$

$2(x(x-5)+12(x-5))=0$

$2(x-5)(x+12)=0$

Set Factors to equal 0:
$x-5=0$

$x=5$

or

$x+12=0$

$x=-12$

However, since we are solving for the side length, the only possible answer is 5 (a shape can't have a side with a negative length.)

The Solution Is: 
$\boxed{x=5}$

I hope this helped!
-Benjamin