Question

Solve each equation by factoring or by using the quadratic formula. The quadratic formula is:

If $a{x}^2+bx+c=0$, with $a\ne 0$, then$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

$z^2=4(2z-3)$

Short answer

The values of z are$z=6$and $z=2$.

Step-by-step solution

- Understand the concept used 

If$a{x}^2+bx+c=0$ , with $a\ne 0$, then $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$. Solving using factoring can be done by splitting the middle term of equation and then making groups and equating them to zero.

- Substitute the values 

It is given that$z^2=4(2z-3)$which can also be written as$z^2-8z+12=0$ .Using quadratic formula$z=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, substitute 1 for a, -8 for b and 12 for c.

Put the values in the formula and get$z=\frac{-\left(-8\right)\pm \sqrt{{\left(-8\right)}^2-4\times \left(1\right)\times \left(12\right)}}{2\left(1\right)}$ .

- Simplify for z 

Simplify the expression for z.

$\begin{array}{c}z=\frac{-\left(-8\right)\pm \sqrt{{\left(-8\right)}^2-4\times \left(1\right)\times \left(12\right)}}{2\left(1\right)}\\ =\frac{8\pm \sqrt{64-48}}{2}\\ =\frac{8\pm \sqrt{16}}{2}\\ =\frac{8\pm 4}{2}\end{array}$

Therefore, either $z=\frac{8+4}{2}=\frac{12}{2}$or $z=\frac{8-4}{2}=\frac{4}{2}$that is., either$z=6$ or$z=2$ .