Quadratic Formula and Completing the Square [H]

The Quadratic Formula and Completing the Square

When a quadratic does not factorise neatly over the integers, two methods are available: completing the square and the quadratic formula. Both always work — for any \( ax^2+bx+c=0 \) with real coefficients. The quadratic formula is derived directly from completing the square on the general form.

The Quadratic Formula

For \( ax^2 + bx + c = 0 \):

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

The \( \pm \) gives the two solutions. The expression under the root, \( b^2-4ac \), is the discriminant. If it is negative, there are no real solutions.

Before applying the formula: always write the equation in the form \(ax^2+bx+c=0\) and identify \(a\), \(b\) and \(c\) explicitly. Sign errors in identifying \(b\) or \(c\) are the most common source of mistakes.

Completing the Square to Solve

Rewrite in the form \( a(x+p)^2+q=0 \), isolate the squared term, then take the square root of both sides.

\[ x^2-6x+7=0 \implies (x-3)^2-9+7=0 \implies (x-3)^2=2 \implies x=3\pm\sqrt{2} \]

Worked Examples

Solve \( 2x^2 + 5x - 4 = 0 \), giving answers to 2 d.p.

\(a=2, b=5, c=-4\). \( \Delta = 25+32=57 \).

\[ x = \frac{-5 \pm \sqrt{57}}{4} \implies x \approx \frac{-5+7.550}{4} \approx 0.64 \text{ or } x \approx \frac{-5-7.550}{4} \approx -3.14 \]
Solve \( x^2 - 4x - 1 = 0 \), leaving answers in surd form.
\[ x = \frac{4 \pm \sqrt{16+4}}{2} = \frac{4 \pm \sqrt{20}}{2} = \frac{4 \pm 2\sqrt{5}}{2} = 2 \pm \sqrt{5} \]
Solve \( x^2 + 8x + 3 = 0 \) by completing the square. Leave in surd form.
\[ (x+4)^2 - 16 + 3 = 0 \implies (x+4)^2 = 13 \implies x = -4 \pm \sqrt{13} \]
Solve \( 2x^2 - 12x + 7 = 0 \) by completing the square.
\[ 2(x^2-6x)+7=0 \implies 2[(x-3)^2-9]+7=0 \implies 2(x-3)^2-18+7=0 \] \[ 2(x-3)^2=11 \implies (x-3)^2=\frac{11}{2} \implies x=3\pm\sqrt{\frac{11}{2}} \]
Solve \( 3x^2 = 7x - 2 \). Give answers to 3 s.f.

Rearrange: \( 3x^2 - 7x + 2 = 0 \). \(a=3, b=-7, c=2\). \( \Delta = 49-24=25 \).

\[ x = \frac{7 \pm 5}{6} \implies x=2 \text{ or } x=\frac{1}{3} \]

(Discriminant is a perfect square here, so it also factorises as \((3x-1)(x-2)=0\).)

Solve \( (x+1)^2 = 3x + 7 \).
\[ x^2+2x+1=3x+7 \implies x^2-x-6=0 \implies (x-3)(x+2)=0 \implies x=3 \text{ or } x=-2 \]

 Key Takeaways

  • Quadratic formula: \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \). Works for all quadratics. Rearrange to \( ax^2+bx+c=0 \) first.
  • If the discriminant \( b^2-4ac < 0 \): no real solutions. If \( = 0 \): one repeated solution. If \( > 0 \): two distinct solutions.
  • Completing the square: rewrite as \( (x+p)^2=k \), then \( x=-p\pm\sqrt{k} \). Useful for exact (surd) answers.
  • When the question asks for a surd form or exact answer, the quadratic formula or completing the square must be used — do not use a decimal approximation.
  • Always check the answer makes sense in context (e.g. a length cannot be negative).