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.

Answer: no real 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 $$${2}{{c}}^{{2}}-{3}{c}=-{21}$$$.

Answer: no real roots.

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 $$$\frac{{-{15}+\sqrt{{{309}}}}}{{2}}$$$.