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