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

$x^2-5x+3=0$

Short answer

The values of x are$x=\frac{5+\sqrt{13}}{2}$and$x=\frac{5-\sqrt{13}}{2}$.

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$x^2-5x+3=0$.Using quadratic formula$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, substitute 1 for a, -5 for b and 3 for c.

Put the values in the formula and get$x=\frac{-\left(-5\right)\pm \sqrt{{\left(-5\right)}^2-4\times \left(1\right)\times \left(3\right)}}{2\left(1\right)}$.

- Simplify for x 

Simplify the expression for x.

$\begin{array}{c}x=\frac{-\left(-5\right)\pm \sqrt{{\left(-5\right)}^2-4\times \left(1\right)\times \left(3\right)}}{2\left(1\right)}\\ =\frac{5\pm \sqrt{25-12}}{2}\\ =\frac{5\pm \sqrt{13}}{2}\end{array}$

Therefore, $x=\frac{5+\sqrt{13}}{2}$and $x=\frac{5-\sqrt{13}}{2}$.