Given:
The three sides of a triangle are:
[tex](x-8),(4x^2-3),(4x^2+4)[/tex]
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,
[tex]Perimeter=(x-8)+(4x^2-3)+(4x^2+4)[/tex]
[tex]Perimeter=8x^2+x-7[/tex]
The perimeter of the triangle 650 units.
[tex]8x^2+x-7=650[/tex]
[tex]8x^2+x-7-650=0[/tex]
[tex]8x^2+x-657=0[/tex]
Splitting the middle term, we get
[tex]8x^2+73x-72x-657=0[/tex]
[tex]x(8x+73)-9(8x+73)=0[/tex]
[tex](x-9)(8x+73)=0[/tex]
Using zero product property, we get
[tex](x-9)=0\text{ and }(8x+73)=0[/tex]
[tex]x=9\text{ and }x=-\dfrac{73}{8}[/tex]
For negative value of x, the side [tex](x-8)[/tex] will be negative, which is not possible because the side of a triangle cannot be negative.
Therefore, the only value of x is 9.
