# Quadratic Equation Formula and the Discriminant

## Related calculator: Quadratic Equation Calculator

Quadratic Equation Formula can be derived from the steps for completing the square (actually, this formula is a general case).

Let's see how to do it.

1. Start from the equation ${a}{{x}}^{{2}}+{b}{x}+{c}={0}$.
2. Divide both sides by ${a}$: ${{x}}^{{2}}+\frac{{b}}{{a}}{x}+\frac{{c}}{{a}}={0}$.
3. Move constant term to the right: ${{x}}^{{2}}+\frac{{b}}{{a}}{x}=-\frac{{c}}{{a}}$.
4. Add ${{\left(\frac{{b}}{{2}}{a}\right)}}^{{2}}=\frac{{{b}}^{{2}}}{{{4}{{a}}^{{2}}}}$ to both sides of the equation: ${{x}}^{{2}}+\frac{{b}}{{a}}{x}+\frac{{{b}}^{{2}}}{{{4}{{a}}^{{2}}}}=-\frac{{c}}{{a}}+\frac{{{b}}^{{2}}}{{{4}{{a}}^{{2}}}}$.
5. Rewrite left hand side: ${{\left({x}+\frac{{b}}{{{2}{a}}}\right)}}^{{2}}=-\frac{{c}}{{a}}+\frac{{{b}}^{{2}}}{{{4}{{a}}^{{2}}}}$.
6. Simplify right hand side: $-\frac{{c}}{{a}}+\frac{{{b}}^{{2}}}{{{4}{{a}}^{{2}}}}=\frac{{-{c}\cdot{\color{red}{{{4}{a}}}}}}{{{a}\cdot{\color{red}{{{4}{a}}}}}}+\frac{{{b}}^{{2}}}{{{4}{{a}}^{{2}}}}=\frac{{-{4}{a}{c}}}{{{4}{{a}}^{{2}}}}+\frac{{{b}}^{{2}}}{{{4}{{a}}^{{2}}}}=\frac{{{{b}}^{{2}}-{4}{a}{c}}}{{{4}{{a}}^{{2}}}}$.
7. Write the final equation: ${{\left({x}+\frac{{b}}{{{2}{a}}}\right)}}^{{2}}=\frac{{{{b}}^{{2}}-{4}{a}{c}}}{{{4}{{a}}^{{2}}}}$.
8. Solve the equation: ${x}+\frac{{b}}{{{2}{a}}}=\sqrt{{\frac{{{{b}}^{{2}}-{4}{a}{c}}}{{{4}{{a}}^{{2}}}}}}$ or ${x}+\frac{{b}}{{{2}{a}}}=-\sqrt{{\frac{{{{b}}^{{2}}-{4}{a}{c}}}{{{4}{{a}}^{{2}}}}}}$.
9. Above equations have roots ${x}_{{1}}=\frac{{-{b}+\sqrt{{{{b}}^{{2}}-{4}{a}{c}}}}}{{{2}{a}}}$ and ${x}_{{2}}=\frac{{-{b}-\sqrt{{{{b}}^{{2}}-{4}{a}{c}}}}}{{{2}{a}}}$.
10. We can write it even more compactly: ${x}_{{{1},{2}}}=\frac{{-{b}\pm\sqrt{{{{b}}^{{2}}-{4}{a}{c}}}}}{{{2}{a}}}$.

Expression ${{b}}^{{2}}-{4}{a}{c}$ is called the discriminant of the quadratic equation.

Since it is under square root, then, depending on its value, quadratic equation will have 0, 1 or 2 roots.

Indeed, if ${{b}}^{{2}}-{4}{a}{c}>{0}$, then $\sqrt{{{{b}}^{{2}}-{4}{a}{c}}}$ exists and we will get two solutions.

If ${{b}}^{{2}}-{4}{a}{c}={0}$, then $\pm\sqrt{{{{b}}^{{2}}-{4}{a}{c}}}=\pm\sqrt{{{0}}}={0}$ and we will get one solution.

If ${{b}}^{{2}}-{4}{a}{c}<{0}$, then there are no real roots, because square root of negative number doesn't exist.

Quadratic Equation Formula: ${\color{brown}{{{a}{{x}}^{{2}}+{b}{x}+{c}={0}}}}$ has following roots: ${\color{ma\genta}{{{x}_{{{1},{2}}}=\frac{{-{b}\pm\sqrt{{{{b}}^{{2}}-{4}{a}{c}}}}}{{{2}{a}}}}}}$.

${\color{green}{{{D}={{b}}^{{2}}-{4}{a}{c}}}}$ is called the Discriminant of the quadratic equation.

• If ${{b}}^{{2}}-{4}{a}{c}>{0}$, equation has 2 real roots.
• If ${{b}}^{{2}}-{4}{a}{c}={0}$, equation has 1 real root.
• If ${{b}}^{{2}}-{4}{a}{c}<{0}$, equation has no real roots.

Let's go through all three cases.

Example 1. Solve the equation using quadratic equation formula: ${5}{{x}}^{{2}}-{9}{x}-{2}={0}$.

Here ${a}={5}$, ${b}=-{9}$, ${c}=-{2}$.

Calculate discriminant: ${D}={{\left(-{9}\right)}}^{{2}}-{4}\cdot{5}\cdot{\left(-{2}\right)}={81}-{\left(-{40}\right)}={121}$.

Since discriminant is positive, then we expect 2 real roots.

${x}_{{1}}=\frac{{-{\left(-{9}\right)}+\sqrt{{{121}}}}}{{{2}\cdot{5}}}=\frac{{{9}+{11}}}{{10}}=\frac{{20}}{{10}}={2}$.

${x}_{{2}}=\frac{{-{\left(-{9}\right)}-\sqrt{{{121}}}}}{{{2}\cdot{5}}}=\frac{{{9}-{11}}}{{10}}=-\frac{{2}}{{10}}=-\frac{{1}}{{5}}$.

Answer: roots are ${2}$ and $-\frac{{1}}{{5}}$.

However, we will not always get rational roots. Roots can also be irrational.

Example 2. Solve the following equation: ${{y}}^{{2}}-{6}{y}+{2}={0}$.

Here ${a}={1}$, ${b}={6}$, ${c}={2}$.

Calculate discriminant: ${D}={{6}}^{{2}}-{4}\cdot{1}\cdot{2}={36}-{8}={28}$.

Since discriminant is positive, then we expect 2 real roots.

${x}_{{1}}=\frac{{-{6}+\sqrt{{{28}}}}}{{{2}\cdot{1}}}=\frac{{-{6}+\sqrt{{{4}\cdot{7}}}}}{{2}}=\frac{{-{6}+\sqrt{{{4}}}\cdot\sqrt{{{7}}}}}{{2}}=\frac{{-{6}+{2}\sqrt{{{7}}}}}{{2}}=-\frac{{6}}{{2}}+\frac{{{2}\sqrt{{{7}}}}}{{2}}=-{3}+\sqrt{{{7}}}$.

${x}_{{2}}=\frac{{-{6}-\sqrt{{{28}}}}}{{{2}\cdot{1}}}=\frac{{-{6}-\sqrt{{{4}\cdot{7}}}}}{{2}}=\frac{{-{6}-\sqrt{{{4}}}\cdot\sqrt{{{7}}}}}{{2}}=\frac{{-{6}-{2}\sqrt{{{7}}}}}{{2}}=-{3}-\sqrt{{{7}}}$.

Answer: roots are $-{3}+\sqrt{{{7}}}$ and $-{3}-\sqrt{{{7}}}$.

Next case is when there is only one root.

Example 3. Solve ${9}{{x}}^{{2}}=-{6}{x}-{1}$.

First of all, write equation in standard form: ${9}{{x}}^{{2}}+{6}{x}+{1}={0}$.

Here ${a}={9}$, ${b}={6}$, ${c}={1}$.

Calculate discriminant: ${D}={{6}}^{{2}}-{4}\cdot{9}\cdot{1}={0}$.

Since discriminant is 0, then we expect 1 real root.

${x}=\frac{{-{6}+\sqrt{{{0}}}}}{{{2}\cdot{9}}}=-\frac{{6}}{{18}}=-\frac{{1}}{{3}}$.

Answer: root is $-\frac{{1}}{{3}}$.

Finally, consider a case, when there are no real roots.

Example 4. Solve the equation, using quadratic equation formula: ${9}{{c}}^{{2}}+{6}{c}=-{2}$.

First of all, write equation in standard form: ${9}{{c}}^{{2}}+{6}{c}+{2}={0}$.

Here ${a}={9}$, ${b}={6}$, ${c}={2}$.

Calculate discriminant: ${D}={{6}}^{{2}}-{4}\cdot{9}\cdot{2}=-{36}$.

Since discriminant is negative, then there are no roots.

Quadratic equation can be used to solve incomplete quadratic equations (although, there are faster methods).

Example 5. Solve incomplete quadratic equation $-{3}{{x}}^{{2}}+{5}={0}$.

Here ${a}=-{3}$, ${b}={0}$, ${c}={5}$.

Calculate discriminant: ${D}={{0}}^{{2}}-{4}\cdot{\left(-{3}\right)}\cdot{5}={60}$.

${x}_{{1}}=\frac{{-{0}+\sqrt{{{60}}}}}{{{2}\cdot{\left(-{3}\right)}}}=-\frac{\sqrt{{{60}}}}{{6}}=-\frac{\sqrt{{{60}}}}{\sqrt{{{36}}}}=-\sqrt{{\frac{{60}}{{36}}}}=-\sqrt{{\frac{{5}}{{3}}}}$.

${x}_{{2}}=\frac{{-{0}-\sqrt{{{60}}}}}{{{2}\cdot{\left(-{3}\right)}}}=\sqrt{{\frac{{5}}{{3}}}}$.

Answer: roots are $-\sqrt{{\frac{{5}}{{3}}}}$ and $\sqrt{{\frac{{5}}{{3}}}}$.

Now, it is time to exercise.

Exercise 1. Solve the following incomplete equation, using qudratic formula: ${2}{{x}}^{{2}}-\frac{{5}}{{3}}{x}={0}$.

Answer: ${0}$ and $\frac{{5}}{{6}}$.

Exercise 2. Solve equation ${24}{{x}}^{{2}}=-{7}{x}+{5}$.

Answer: $-\frac{{5}}{{8}}$ and $\frac{{1}}{{3}}$.

Exercise 3. Find roots of the equation ${2}{{c}}^{{2}}-{3}{c}=-{21}$.

Exercise 4. Solve $\frac{{1}}{{4}}{{x}}^{{2}}-{5}{x}+{25}={0}$.
Answer: ${10}$.
Exercise 5. Solve the following equation, using the quadratic formula: $\frac{{1}}{{3}}{{y}}^{{2}}+{5}{y}-{7}={0}$.
Answer: $\frac{{-{15}-\sqrt{{{309}}}}}{{2}}$ and $\frac{{-{15}+\sqrt{{{309}}}}}{{2}}$.