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

$y^2=13y$

Short answer

The values of y are$y=13$and.$y=0$

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$y^2=13y$, which can also be written as$y^2-13y=0$ .Using quadratic formula$y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, substitute 1 for a, -13 for b and 0 for c.

Put the values in the formula and get$y=\frac{-(-13)\pm \sqrt{{(-13)}^2-4\times \left(1\right)\times \left(0\right)}}{2\left(1\right)}$ .

- Simplify for y 

Simplify the expression for y.

$\begin{array}{c}y=\frac{-(-13)\pm \sqrt{{(-13)}^2-4\times \left(1\right)\times \left(0\right)}}{2\left(1\right)}\\ =\frac{13\pm \sqrt{169-0}}{2}\\ =\frac{13\pm \sqrt{169}}{2}\\ =\frac{13\pm 13}{2}\end{array}$

Therefore, either $y=\frac{13+13}{2}=\frac{26}{2}$or$y=\frac{13-13}{2}=\frac{0}{2}$ that is., either $y=13$or $y=0$.