Solving Quadratic equations by completing the squares - SS2 Mathematics Lesson Note

Some quadratic equations cannot be solved by factorisation. These equations can be solved via the completing the square method which involving making the equation a perfect square that can be factorised.

Example 1 Solve the equation \({2x}^{2} + 3x - 2 = 0\) using the completing the square method.

Solution

STEP 1: make the coefficient of \(x^{2}\) unity by dividing every term by \(2\)

\(x^{2} + \frac{3}{2}x - 1 = 0\)

STEP 2: transfer any constant term to the RHS of the equality sign

\(x^{2} + \frac{3}{2}x = 1\)

STEP 3: make the LHS a perfect square by adding the square of half the coefficient of \(x\) to both sides

\(x^{2} + \frac{3}{2}x + {(\frac{3}{4})}^{2} = 1 + {(\frac{3}{4})}^{2}\)

\(x^{2} + \frac{3}{2}x + \frac{9}{16} = 1 + \frac{9}{16}\)

STEP 4: factorise the LHS

\({(x + \frac{3}{4})}^{2} = \frac{25}{16}\)

\(x + \frac{3}{4} = \ \sqrt{\frac{25}{16}}\)

\(x + \frac{3}{4} = \ \pm \frac{5}{4}\)

\(x = - \frac{3}{4} \pm \frac{5}{4} = \ - \frac{3}{4} + \frac{5}{4}\ or\ - \frac{3}{4} - \frac{5}{4}\)

\(x = \frac{2}{4}\ or - \frac{8}{4}\)

\(x = \frac{1}{2}\ or - 2\)

From this method, we obtain the much easier to use quadratic formula: \(x = \frac{- b \pm \sqrt{b^{2} - 4ac}}{2a}\). Here, \(b^{2} - 4ac\) is called the discriminant denoted in math by the letter D and is used to determine what nature the roots possess. Given an equation, \({ax}^{2} + bx + c = 0\):

1. If \(b^{2} - 4ac > 0\), then the roots are real and distinct

If \(b^{2} - 4ac < 0\), then the roots are imaginary or complex (usually involving the square of a negative number)

If \(b^{2} - 4ac = 0\), then the roots are real and equal (coincidental)