Chapter 9.6, Problem 52HP

To determine

The number of real zeros of equation $f\left(x\right)=x^2-8x+16$ and find the number of real solutions.

Answer to Problem 52HP

The factor of the given equation is $x=4,4$

  $D=0\text{ there are two same real roots}$

Explanation of Solution

Given information:

The given equation, $f\left(x\right)=x^2-8x+16$

Concept used:

• The discriminant is given by, $D=b^2-4ac$
• For a quadratic equation $a{x}^2+bx+c=0$ the number of roots depend on:
• If discriminant is negative, $b^2-4ac<0$, then there is no real roots
• If discriminant is zero, $b^2-4ac=0$, then there are two same real roots
• If discriminant is positive, $b^2-4ac>0$, then there are two distinct real roots

Calculation:

To find the factors of the given equation $f\left(x\right)=x^2-8x+16$ we find factor as below

  $\begin{array}{l}{x}^2-8x+16=0\\ x^2-\left(4+4\right)x+16=0\\ x^2-4x-4x+16=0\\ x\left(x-4\right)-4\left(x-4\right)=0\\ \left(x-4\right)\left(x-4\right)=0\\ \Rightarrow x=4,4\end{array}$

Thus, the factor of the given equation is

  $x=4,4$

To find the discriminant of the given quadratic equation first rewrite the given equation in standard form So it gives

  $x^2-8x+16=0$

Now compare this equation with the general equation,

  $a{x}^2+bx+c=0$ to get the value of the variables, $a=1,b=-8,c=16$ . So, the discriminant of the given equation is

  $\begin{array}{c}D=b^2-4ac\\ ={\left(-8\right)}^2-4\left(1\right)\left(16\right)\\ =64-64\\ =0\end{array}$

So, the discriminant of the given equation is $0$ .

Since the discriminant is zero, so there are there are two same real roots of the given equation.