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

$y^2-6y+9=0$

Short answer

The values of y are $y=3$and $y=3$.

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 $y^2-6y+9=0$.Using quadratic formula $y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, substitute 1 for a, -6 for b and 9 for c.

Put the values in the formula and get $y=\frac{-\left(-6\right)\pm \sqrt{{\left(-6\right)}^2-4\times \left(1\right)\times \left(9\right)}}{2\left(1\right)}$.

- Simplify for y 

Simplify the expression for y.

$\begin{array}{c}y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{-\left(-6\right)\pm \sqrt{{\left(-6\right)}^2-4\left(1\right)\left(9\right)}}{2\left(1\right)}\\ =\frac{6\pm \sqrt{36-36}}{2}\\ =\frac{6\pm 0}{2}\end{array}$

Therefore, either $y=\frac{6+0}{2}=\frac{6}{2}$ or $y=\frac{6-0}{2}=\frac{6}{2}$ that is., either $y=3$ or $y=3$.