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

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:

[tex]H=\sqrt{Leg_{1}^2+Leg_{2}^2} \\ \\ \therefore (\sqrt{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[/tex]

By using the quadratic formula:

[tex]a=2 \\ b=2 \\ c=-60 \\ \\ x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ \\ x=\frac{-2 \pm \sqrt{2^2-4(2)(-60)}}{2(2)} \\ \\ x_{1}=5 \ and \ x_{2}=-6[/tex]

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

[tex]\boxed{x=5}[/tex]

freckledspots freckledspots · Answer 2 · 2019-10-04T22:36:15+02:00

Answer:

5 is the value for x.

Step-by-step explanation:

By Pythagorean Theorem we have:

[tex](\sqrt{61})^2=(x)^2+(x+1)^2[/tex]

[tex]61=x^2+(x+1)^2[/tex]

Expand the binomial square using [tex](x+y)^2=x^2+2xy+y^2)[/tex]:

[tex]61=x^2+x^2+2x+1[/tex]

Combine like terms:

[tex]61=2x^2+2x+1[/tex]

Subtract 61 on both sides:

[tex]2x^2+2x-60=0[/tex]

Divide both sides by 2:

[tex]x^2+x-30=0[/tex]

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

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

[tex](x+6)(x-5)=0[/tex]

This implies:

[tex]x+6=0[/tex] or [tex]x-5=0[/tex]

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

[tex]x=-6[/tex]

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

[tex]x=5[/tex]

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

Let's check:

[tex]5^2+(5+1)^2[/tex]

[tex]5^2+(6)^2[/tex]

[tex]25+36[/tex]

[tex]61[/tex]

[tex](\sqrt{61})^2[/tex]