Chapter 4, Problem 26AR

To determine

To solve the given equation, $z^2=4\left(2z-3\right)$ by factoring or quadratic formula.

Answer to Problem 26AR

  $x=6,2$

Explanation of Solution

Given:

The equation is, $z^2=4\left(2z-3\right)$

Calculation:

We can solve the given equation by using quadratic formula.

That is,

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Write the given equation as

  $\begin{array}{l}{z}^2=4\left(2z-3\right)\\ z^2=8z-12\\ z^2-8z+12=0\end{array}$

The equation $z^2-8z+12=0$ , which is in the form of $a{x}^2+bx+c=0$ . Where, $a=1,b=-8\text{ and c =12}$

Now, Substitute the values in the formula then we get that,

  $\begin{array}{l}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ x=\frac{-\left(-8\right)\pm \sqrt{{\left(-8\right)}^2-4\left(1\right)\left(12\right)}}{2\left(1\right)}\\ x=\frac{8\pm \sqrt{64-48}}{2}\\ x=\frac{8\pm \sqrt{16}}{2}\\ x=\frac{8\pm 4}{2}\\ x=\frac{8+4}{2},\frac{8-4}{2}\\ x=\frac{12}{2},\frac{4}{2}\\ x=6,2\end{array}$

Therefore, the roots for the given equation, $z^2=4\left(2z-3\right)$ are $x=6,2$ .