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

$x^2=3x+4$

Short answer

The values of x are $x=4$and $x=-1$.

Step-by-step solution

Step 1.  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.

Step 2.  Substitute the values 

It is given that $x^2=3x+4$which can also be written as $x^2-3x-4=0$.Using quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, substitute 1 for a, -3 for b and -4 for c.

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

Step 3.  Simplify for x 

Simplify the expression for x.

$\begin{array}{c}x=\frac{-\left(-3\right)\pm \sqrt{{\left(-3\right)}^2-4\times \left(1\right)\times \left(-4\right)}}{2\left(1\right)}\\ =\frac{3\pm \sqrt{9+16}}{2}\\ =\frac{3\pm \sqrt{25}}{2}\\ =\frac{3\pm 5}{2}\end{array}$

Therefore, either $x=\frac{3+5}{2}=\frac{8}{2}$ or $x=\frac{3-5}{2}=\frac{-2}{2}$ that is., either $x=4$ or $x=-1$.