See also: Quadratic Functions and Quadratic Equations, Completing Square, Quadratic Equation Factorisation, numbers

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Given a quadratic equation of the following form,

$$a<!--\; InvisibleTimes\; -->x^2+b<!--\; InvisibleTimes\; -->x+c=0$$

the roots can be found using the formula below.

$$x=\frac{-b\pm \sqrt{b^2-4<!--\; InvisibleTimes\; -->a<!--\; InvisibleTimes\; -->c}}{2<!--\; InvisibleTimes\; -->a}$$

Using the above formula is probably the easiest and most straightforward way to solve or find the roots of a quadratic equation.

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Examples:

1. Find the roots of the quadratic equation

   $$x^2-4<!--\; InvisibleTimes\; -->x+3=0$$

   We subsitute the values of the coefficients $a$, $b$ and $c$ to the formula to find $x$.

   In this case, $a=1$, $b=-4$ and $c=3$. So

   $$\begin{array}{rl}x& =\frac{-(-4)\pm \sqrt{{(-4)}^2-4\cdot 1\cdot 3}}{2\cdot 1}\\ x& =\frac{4\pm \sqrt{16-12}}{2}\\ x& =\frac{4\pm \sqrt{4}}{2}\\ x& =\frac{4\pm 2}{2}\\ x_1& =\frac{4-2}{2}=1\\ x_2& =\frac{4+2}{2}=3\end{array}$$

2. Find the roots of the quadratic equation

   $$x^2-6<!--\; InvisibleTimes\; -->x+9=0$$

   Similar to the first example, we subsitute the values of the coefficients $a$, $b$ and $c$ to the formula to find $x$.

   In this case, $a=1$, $b=-6$ and $c=9$. So

   $$\begin{array}{rl}x& =\frac{-(-6)\pm \sqrt{{(-6)}^2-4\cdot 1\cdot 9}}{2\cdot 1}\\ x& =\frac{6\pm \sqrt{36-36}}{2}\\ x& =\frac{6\pm \sqrt{0}}{2}\\ x& =\frac{6\pm 0}{2}\\ x_{1\text{,}2}& =\frac{6}{2}\\ x_{1\text{,}2}& =3\end{array}$$

   This quadratic equation has only 1 root as expected because the discriminant ( $b^2-4<!--\; InvisibleTimes\; -->a<!--\; InvisibleTimes\; -->c$ ) is $0$.

3. Find the roots of the quadratic equation

   $$2<!--\; InvisibleTimes\; -->x^2+2<!--\; InvisibleTimes\; -->x+5=0$$

   Similar to the above examples, we subsitute the values of the coefficients $a$, $b$ and $c$ to the formula to find $x$.

   In this case, $a=2$, $b=2$ and $c=5$. So

   $$\begin{array}{rl}x& =\frac{-2\pm \sqrt{2^2-4\cdot 2\cdot 5}}{2\cdot 2}\\ x& =\frac{-2\pm \sqrt{4-40}}{4}\\ x& =\frac{-2\pm \sqrt{-36}}{4}\\ x& =\frac{-2\pm 6<!--\; InvisibleTimes\; -->\sqrt{-1}}{4}\\ x& =\frac{-2\pm 6<!--\; InvisibleTimes\; -->i}{4}\\ x& =\frac{-1\pm 3<!--\; InvisibleTimes\; -->i}{2}\\ x_1& =\frac{-1-3<!--\; InvisibleTimes\; -->i}{2}\\ x_2& =\frac{-1+3<!--\; InvisibleTimes\; -->i}{2}\end{array}$$

   This quadratic equation has 2 complex roots as expected because the discriminant ( $b^2-4<!--\; InvisibleTimes\; -->a<!--\; InvisibleTimes\; -->c$ ) is negative.

Confused and have questions? We’ve got answers. With Chegg Study, you can get step-by-step solutions to your questions from an expert in the field.

By Jimmy Sie

See also: Quadratic Functions and Quadratic Equations, Completing Square, Quadratic Equation Factorisation, numbers