Chapter 4, Problem 25AR

To determine

To solve the given equation, ${\left(y-5\right)}^2=16$ by factoring or quadratic formula.

Answer to Problem 25AR

  $x=9,1$

Explanation of Solution

Given:

The equation is, ${\left(y-5\right)}^2=16$

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

  $\begin{array}{l}{\left(y-5\right)}^2=16\\ \left(y^2+25-10y\right)=16\\ y^2+25-10y-16=0\\ y^2-10y+9=0\end{array}$

The equation $y^2-10y+9=0$ , which is in the form of $a{x}^2+bx+c=0$ . Where, $a=1,b=-10\text{ and c =9}$

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

  $\begin{array}{l}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ x=\frac{-\left(-10\right)\pm \sqrt{{\left(-10\right)}^2-4\left(1\right)\left(9\right)}}{2\left(1\right)}\\ x=\frac{10\pm \sqrt{100-36}}{2}\\ x=\frac{10\pm \sqrt{64}}{2}\\ x=\frac{10\pm 8}{2}\\ x=\frac{10+8}{2},\frac{10-8}{2}\\ x=\frac{18}{2},\frac{2}{2}\\ x=9,1\end{array}$

Therefore, the roots for the given equation, ${\left(y-5\right)}^2=16$ are $x=9,1$ .