Question

A triangle has side lengths of

(x-8) (4x2 – 3) and (4x2 + 4).
If the perimeter is 650 units, what is the value of x?​

Answer 1

Given:

The three sides of a triangle are:

`(x-8),(4x^2-3),(4x^2+4)`

The perimeter of the triangle 650 units.

To find:

The value of x.

Solution:

Perimeter of a triangle is the sum of all three sides of the triangle. So,

`Perimeter=(x-8)+(4x^2-3)+(4x^2+4)`

`Perimeter=8x^2+x-7`

The perimeter of the triangle 650 units.

`8x^2+x-7=650`

`8x^2+x-7-650=0`

`8x^2+x-657=0`

Splitting the middle term, we get

`8x^2+73x-72x-657=0`

`x(8x+73)-9(8x+73)=0`

`(x-9)(8x+73)=0`

Using zero product property, we get

`(x-9)=0\text{ and }(8x+73)=0`

`x=9\text{ and }x=-(73)/(8)`

For negative value of x, the side
`(x-8)` will be negative, which is not possible because the side of a triangle cannot be negative.

Therefore, the only value of x is 9.