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Algebra

Question 8 of 9

Factorising Quadratic Equations

Question

Factorise \(\dfrac{6}{x^2} = \dfrac{3x + 7}{x}\)

Show solution Hide solution Fully worked — 6 steps
  1. Multiply both sides by \(x^2\)
    \[ \frac{6}{x^2} = \frac{3x + 7}{x} \]
    \[ 6 = (3x + 7)(x) \]
  2. Multiply \((3x + 7)\) and \((x)\)
    \[ 6 = 3x^2 + 7x \]
  3. Rearrange in the form \(ax^2 + bx + c = 0\)
    \[ 3x^2 + 7x - 6 = 0 \]
  4. Factorise
    \[ (3x - 2)(x + 3) = 0 \]
  5. Using the zero-factor theorem, equate each factor to 0

    Then either

    \[ 3x - 2 = 0 \Rightarrow x = \frac{2}{3} \]

    or

    \[ x + 3 = 0 \Rightarrow x = -3 \]
  6. State the solutions

    The solutions are \(x = \frac{2}{3}\) or \(x = -3\). That is, \(x\) is two thirds and \(x\) is minus three.