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How do you solve quadratic equations by factorising, the quadratic formula and completing the square?

Solving quadratic equations by factorising, by the quadratic formula and by completing the square (Higher tier), and interpreting the roots and the turning point of the curve.

A focused answer to the Edexcel GCSE Mathematics algebra content on quadratic equations, covering solving by factorising, the quadratic formula, completing the square at Higher tier, and interpreting the roots and turning point.

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  1. What this dot point is asking
  2. Solving by factorising
  3. The quadratic formula
  4. Completing the square (Higher)
  5. Interpreting roots and the turning point

What this dot point is asking

Edexcel wants you to solve quadratic equations by three methods (factorising, the quadratic formula, and at Higher tier completing the square) and to read the roots as the points where the curve y=ax2+bx+cy = ax^2 + bx + c crosses the xx-axis. Quadratics are one of the highest-value Higher topics: they reappear in graph sketching, simultaneous equations, area problems and rates of change, so fluency here pays off across the whole paper.

Solving by factorising

Factorising is the fastest method when it works, and it is the expected method on non-calculator papers. The key fact is the zero-product principle: if two things multiply to give zero, at least one of them must be zero.

For a quadratic x2+bx+cx^2 + bx + c (where a=1a = 1), find two numbers that multiply to cc and add to bb. To solve x2+5x+6=0x^2 + 5x + 6 = 0, you need two numbers multiplying to 66 and adding to 55: those are 22 and 33, so (x+2)(x+3)=0(x + 2)(x + 3) = 0 and x=−2x = -2 or x=−3x = -3.

When a≠1a \ne 1 you split the middle term. For 6x2+11x−10=06x^2 + 11x - 10 = 0, multiply aa by cc to get −60-60, then find two numbers multiplying to −60-60 and adding to 1111: those are 1515 and −4-4. Rewrite as 6x2+15x−4x−10=06x^2 + 15x - 4x - 10 = 0, factor by grouping into 3x(2x+5)−2(2x+5)=03x(2x + 5) - 2(2x + 5) = 0, so (3x−2)(2x+5)=0(3x - 2)(2x + 5) = 0 and x=23x = \tfrac{2}{3} or x=−52x = -\tfrac{5}{2}.

The quadratic formula

When a quadratic does not factorise over the integers, the formula always works and is the standard tool on calculator papers. It is provided on the Edexcel formulae sheet.

If the discriminant is positive there are two distinct real roots; if it is zero there is one repeated root (the curve touches the xx-axis); if it is negative there are no real roots (the curve never crosses the xx-axis). Examiners often hide a discriminant test inside a "how many solutions" question.

Completing the square (Higher)

Completing the square is the Higher-tier method that both solves the equation and exposes the turning point of the parabola.

For example x2+6x+5=(x+3)2−9+5=(x+3)2−4x^2 + 6x + 5 = (x + 3)^2 - 9 + 5 = (x + 3)^2 - 4, which has a minimum at (−3,−4)(-3, -4). To solve x2+6x+5=0x^2 + 6x + 5 = 0 you then write (x+3)2=4(x + 3)^2 = 4, take square roots to get x+3=±2x + 3 = \pm 2, and finish with x=−1x = -1 or x=−5x = -5.

Interpreting roots and the turning point

The roots are the xx-intercepts of y=ax2+bx+cy = ax^2 + bx + c. A positive aa gives a U-shaped curve with a minimum; a negative aa gives an n-shaped curve with a maximum. The line of symmetry passes through the turning point at x=−b2ax = -\tfrac{b}{2a}, which sits exactly halfway between the two roots. Completing the square is the cleanest route to the turning point, because the bracket gives the xx-coordinate and the constant gives the yy-coordinate directly. This is why an exam may ask you to "write in the form (x+a)2+b(x + a)^2 + b and hence state the minimum point".

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20193 marksSolve x2+2x−15=0x^2 + 2x - 15 = 0 by factorising. (Higher tier, Paper 1, non-calculator.)
Show worked answer →

Find two numbers that multiply to −15-15 and add to +2+2. Those are +5+5 and −3-3.

Factorise: (x+5)(x−3)=0(x + 5)(x - 3) = 0.

Set each bracket to zero: x+5=0x + 5 = 0 gives x=−5x = -5; x−3=0x - 3 = 0 gives x=3x = 3.

Markers award one mark for the factorisation and one for each correct root. A common loss is giving the roots as +5+5 and −3-3 (sign slip) instead of −5-5 and +3+3.

Edexcel 20213 marksSolve 2x2−7x+1=02x^2 - 7x + 1 = 0, giving your answers to 2 decimal places. (Higher tier, Paper 2, calculator.)
Show worked answer →

The expression does not factorise nicely, so use the quadratic formula with a=2a = 2, b=−7b = -7, c=1c = 1.

Discriminant: b2−4ac=(−7)2−4(2)(1)=49−8=41b^2 - 4ac = (-7)^2 - 4(2)(1) = 49 - 8 = 41.

Substitute: x=7±414x = \dfrac{7 \pm \sqrt{41}}{4}.

Since 41≈6.403\sqrt{41} \approx 6.403, the roots are x≈7+6.4034≈3.35x \approx \dfrac{7 + 6.403}{4} \approx 3.35 and x≈7−6.4034≈0.15x \approx \dfrac{7 - 6.403}{4} \approx 0.15.

Markers give method marks for correct substitution and discriminant, then accuracy marks for both rounded roots. Forgetting the ±\pm caps you at part marks.

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