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Quadratic formula

The quadratic formula provides a straightforward way to find the solutions of any quadratic equation, especially when factoring is difficult or impossible. By becoming familiar with this formula, you'll gain a reliable tool for solving quadratic equations and deepen your understanding of algebraic relationships.

Solving quadratic equations

The solutions to any quadratic equation with the form \(ax^{2}+bx+c=0\) can be found by substituting the values \(a\), \(b\) and \(c\) into the quadratic formula:

\[ x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\]

Discriminant

Solutions to quadratic equations may be real or complex numbers. Here, we will only consider real solutions.

Remember from quadratic factorisation that the discriminant (\(\Delta)\) can be used to determine the number and type of solutions.

  • If \(\Delta<0\), there are no (real) solutions. The graph of \(y=ax^{2}+bx+c\) does not touch or cut the \(x\)-axis.
  • If \(\Delta=0\), there is one solution. The graph of \(y=ax^{2}+bx+c\) touches the \(x\)-axis at the value of \(x\).
  • If \(\Delta>0\), there are two solutions. The graph of \(y=ax^{2}+bx+c\) cuts the \(x\)-axis at these values of \(x\).
  • If \(b^{2}-4ac\) is a perfect square, the solutions will be rational. This means that the values of \(x\) can be expressed as a fraction \(\dfrac{p}{q}\), where \(p\) and \(q\neq0\) are integers.

Example 1 – solving quadratic equations

Solve the quadratic equation:
\[x^{2}-5x+4=0\]

Identify \(a\), \(b\) and \(c\).

\[a=1,\,b=-5,\,c=4\]

Substitute into the quadratic formula:

\[\begin{align*} x & = \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\\
& = \frac{-(-5)\pm\sqrt{(-5)^{2}-4(1)(4)}}{2(1)}\\
& = \frac{5\pm\sqrt{25-16}}{2}\\
& = \frac{5\pm\sqrt{9}}{2}\\
& = \frac{5\pm3}{2}\\
& = \frac{5+3}{2}\ \textrm{ or }\ \frac{5-3}{2}\\
& = \frac{8}{2}\ \textrm{ or }\ \frac{2}{2}\\
& = 4\ \textrm{ or }\ 1
\end{align*}\]

The solution is \(x=1\) or \(x=4\).

Solve the quadratic equation:
\[x^{2}+x+10=0\]

Identify \(a\), \(b\) and \(c\).

\[a=1,\,b=1,\,c=10\]

Substitute into the quadratic formula:

\[\begin{align*} x & = \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\\
& = \frac{-1\pm\sqrt{1^{2}-4(1)(10)}}{2(1)}\\
& = \frac{-1\pm\sqrt{1-40}}{2}\\
& = \frac{-1\pm\sqrt{-39}}{2}
\end{align*}\]

Since the \(\sqrt{-39}\) does not exist, the equation has no solution. There is a solution if we use complex numbers but we are looking for real solutions.

We can confirm this by calculating the discriminant:

\[\begin{align*} b^{2}-4ac & = (1)^{2}-4(1)(10)\\
& = 1-40\\
& = -30
\end{align*}\]

\(\Delta<0\) so there are no real solutions.

Your turn – solving quadratic equations

  1. Find the real solutions to the following quadratic equations.
    1. \(x^{2}-4x-7=0\)
    2. \(x^{2}-2x-2=0\)
    3. \(x^{2}+6x-9=0\)
    4. \(x^{2}-x-7=0\)
    5. \(4x^{2}-12x+6=0\)
    6. \(2x^{2}+x-2=0\)
    7. \(x^{2}-4x+2=0\)
    8. \(2x^{2}-3x+2=0\)
    9. \(3x^{2}+5x-7=0\)
    10. \(3x^{2}=x+1\)
    11. \(4x^{2}+x+3=0\)
    12. \(\dfrac{3x+1}{2}=\dfrac{x+1}{x}\)

    1. \(x =\dfrac{4\pm\sqrt{44}}{2} = 2\pm\sqrt{11}\)
    2. \(x = \dfrac{2\pm\sqrt{12}}{2} = 1\pm\sqrt{3}\)
    3. \(x = \dfrac{-6\pm\sqrt{72}}{2} = -3\pm\sqrt{2}\)
    4. \(x = \dfrac{1\pm\sqrt{29}}{2}\)
    5. \(x = \dfrac{12\pm\sqrt{48}}{2} = \dfrac{3\pm\sqrt{3}}{2}\)
    6. \(x = \dfrac{-1\pm\sqrt{17}}{4}\)
    7. \(x = \dfrac{4\pm\sqrt{8}}{2} = 2\pm\sqrt{2}\)
    8. No real solutions
    9. \(x = \dfrac{-5\pm\sqrt{109}}{6}\)
    10. \(x = \dfrac{1\pm\sqrt{13}}{6}\)
    11. No real solutions
    12. \(x = -\dfrac{2}{3}\) or \(x=1\)

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Further resources

Quadratic factorisation

Need to review quadratic factorisation? Use this resource.

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