Question

For what real values of $c$ is $x^2 - 8x + c$ the square of a binomial?

Answer 1

Value of c is 16 for which equation  $f(x) = x^2-8x+c$  is a square of binomial !

Step-by-step explanation:

Here we meed to find the value of c for which equation f(x) = x^2-8x+c or ,  $f(x) = x^2-8x+c$ is a square of a binomial . Let's find out:

We know that

⇒ $(x-a)^2 = x^2-2(a)(x)+a^2$   ..........(1)

Let's simplify given equation in question

⇒ $f(x) = x^2-8x+c$

⇒ $f(x) = x^2-2(4)x+c$

Comparing this equation with (1) we get :

$a=4 , c=a^2$

⇒ $c=a^2$

⇒ $c=4^2$

⇒ $c=16$

Therefore , Value of c is 16 for which equation  $f(x) = x^2-8x+c$  is a square of binomial !