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Algebra

Question 7 of 9

Factorising Quadratic Equations

Question

Factorise \(6m^4 = m^2 + 12\)

Show solution Hide solution Fully worked — 8 steps
  1. Let \(a = m^2\)
    \[ 6m^4 = m^2 + 12 \]
    \[ 6a^2 = a + 12 \]
  2. Rearrange in the form \(ax^2 + bx + c = 0\)
    \[ 6a^2 - a - 12 = 0 \]
  3. Factorise
    \[ (3a + 4)(2a - 3) = 0 \]
  4. Using the zero-factor theorem, equate each factor to 0

    Then either

    \[ 3a + 4 = 0 \quad \text{or} \quad 2a - 3 = 0 \]
    \[ a = \frac{-4}{3} \ \text{ or } \ a = \frac{3}{2} \]
  5. Return to \(m\)

    Since \(a = m^2\), we have:

    \[ m^2 = \frac{-4}{3} \ \text{ or } \ m^2 = \frac{3}{2} \]
  6. Take the square root
    \[ m = \pm\sqrt{\frac{-4}{3}} \ \text{ or } \ m = \pm\sqrt{\frac{3}{2}} \]

    Using \(\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}\):

    \[ m = \pm\frac{\sqrt{-4}}{\sqrt{3}} \ \text{ or } \ m = \pm\frac{\sqrt{3}}{\sqrt{2}} \]
  7. Rationalise the denominators

    Using \(\frac{b}{\sqrt{a}} = \frac{b}{\sqrt{a}} \times \frac{\sqrt{a}}{\sqrt{a}} = \frac{b\sqrt{a}}{a}\), and since \(\sqrt{-1} = i\):

    \[ m = \pm\frac{2i\sqrt{3}}{3} \ \text{ or } \ m = \pm\frac{\sqrt{6}}{2} \]
  8. State the solutions

    Therefore, the solutions are \(m = \pm\frac{2i\sqrt{3}}{3}\) or \(m = \pm\frac{\sqrt{6}}{2}\). That is, \(m\) is plus or minus two \(i\) root three over three, and \(m\) is plus or minus root six over two.