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$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

${(y-5)}^2=16$

Short answer

The values of y are$y=9$and.$y=1$

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${(y-5)}^2=16$which can also be written as$y^2-10y+9=0$ .Using quadratic formula$y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, substitute 1 for a, -10 for b and 9 for c.

Put the values in the formula and get$y=\frac{-\left(-10\right)\pm \sqrt{{\left(-10\right)}^2-4\times \left(1\right)\times \left(9\right)}}{2\left(1\right)}$ .

- Simplify for y 

Simplify the expression for y.

$\begin{array}{c}y=\frac{-\left(-10\right)\pm \sqrt{{\left(-10\right)}^2-4\times \left(1\right)\times \left(9\right)}}{2\left(1\right)}\\ =\frac{10\pm \sqrt{100-36}}{2\left(1\right)}\\ =\frac{10\pm \sqrt{64}}{2}\\ =\frac{10\pm 8}{2}\end{array}$

Therefore, either $y=\frac{10+8}{2}=\frac{18}{2}$or $y=\frac{10-8}{2}=\frac{2}{2}$that is., either $y=9$or$y=1$ .