Question

If 61 is the longest side length in the triangle , find the value of x that makes the triangle above a right triangle.

Answer 1

5 is the value for x.

Explanation:

By Pythagorean Theorem we have:

`(√(61))^2=(x)^2+(x+1)^2`

`61=x^2+(x+1)^2`

Expand the binomial square using
`(x+y)^2=x^2+2xy+y^2)`:

`61=x^2+x^2+2x+1`

Combine like terms:

`61=2x^2+2x+1`

Subtract 61 on both sides:

`2x^2+2x-60=0`

Divide both sides by 2:

`x^2+x-30=0`

Find two numbers that multiply to be -30 and add to be 1.

Those numbers are 6 and -5 so the factored form is:

`(x+6)(x-5)=0`

This implies:

`x+6=0` or
`x-5=0`

Solving the first equation by subtracting 6 on both sides gives:

`x=-6`

Solving the second equation by adding 5 on both sides gives:

`x=5`

So since a side measurement can bot be negative units long, the only answer that x can be is 5.

Let's check:

`5^2+(5+1)^2`

`5^2+(6)^2`

`25+36`

`61`

`(√(61))^2`

Answer 2

`\boxed{x=5}`

Explanation:

A right triangle is a triangle that has a 90° degree side. The hypotenuse is the largest side while the legs are the other two sides. To solve this problem we must use the Pythagorean Theorem as follows:

`H=\sqrt{Leg_(1)^2+Leg_(2)^2} \\ \\ \therefore (√(61))^2=(x+1)^2+x^2 \\ \\ \therefore x^2+2x+1+x^2=61 \\ \\ \therefore 2x^2+2x+1=61\\ \\ \therefore 2x^2+2x-60=0`

By using the quadratic formula:

`a=2 \\ b=2 \\ c=-60 \\ \\ x=(-b \pm √(b^2-4ac))/(2a) \\ \\ x=(-2 \pm √(2^2-4(2)(-60)))/(2(2)) \\ \\ x_(1)=5 \ and \ x_(2)=-6`

Since lengths only admit positive number, we only have one solution which is:

`\boxed{x=5}`