Quadratic Equations

A quadratic equation is one in which the highest power of the unknown is $2$2 — its general form is $ax^{2} + bx + c = 0$ax2+bx+c=0. Such equations usually have two solutions (called roots), one solution, or none at all. This lesson covers solving by factorising, solving with the quadratic formula, and — at Higher tier — completing the square, together with what the roots actually mean in a real context. Carrying out each method is AO1; interpreting roots in a worded problem and deciding which solution is valid is AO3, and explaining your reasoning is AO2.

Key Vocabulary

Term	Meaning
Quadratic equation	An equation of the form $ax^{2} + bx + c = 0$ax2+bx+c=0 with $a \ne 0$a=0
Root (solution)	A value of the unknown that satisfies the equation
Factorise	Write the quadratic as a product of two brackets
Quadratic formula	$x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$x=2a−b±b2−4ac​​
Discriminant	The quantity $b^{2} - 4ac$b2−4ac, which tells you how many roots there are
Complete the square	Rewrite $x^{2} + bx + c$x2+bx+c as $(x + p)^{2} + q$(x+p)2+q
Exact value	A value left as a surd or fraction, not rounded

Before diving into methods, it helps to picture what a quadratic equation represents. The graph of $y = ax^{2} + bx + c$y=ax2+bx+c is a smooth U-shaped (or n-shaped) curve called a parabola. Solving $ax^{2} + bx + c = 0$ax2+bx+c=0 means finding where that curve crosses the $x$x-axis. A parabola can cross the axis twice (two roots), touch it once (one repeated root), or miss it entirely (no real roots) — which is exactly why a quadratic can have two, one, or no solutions. Keeping this picture in mind makes the number of answers feel natural rather than surprising.

Solving by Factorising

If a product equals zero, at least one of the factors must be zero — this is called the zero-product property, and it is the engine behind this method. So once a quadratic is factorised and set equal to zero, set each bracket to zero in turn. The equation must be arranged with all terms on one side and zero on the other before you factorise; this is the step students most often skip.

Worked Example 1

Solve $x^{2} + 5x + 6 = 0$x2+5x+6=0.

Factorise: two numbers multiplying to $6$6 and adding to $5$5 are $2$2 and $3$3, so $(x + 2)(x + 3) = 0$(x+2)(x+3)=0. Then $x + 2 = 0$x+2=0 or $x + 3 = 0$x+3=0.

Answer: $x = -2$x=−2 or $x = -3$x=−3

Worked Example 2

Solve $x^{2} - 7x + 10 = 0$x2−7x+10=0.

Two numbers multiplying to $10$10 and adding to $-7$−7 are $-2$−2 and $-5$−5, so $(x - 2)(x - 5) = 0$(x−2)(x−5)=0.

Answer: $x = 2$x=2 or $x = 5$x=5

Worked Example 3

Solve $x^{2} - 9 = 0$x2−9=0.

This is a difference of two squares: $(x + 3)(x - 3) = 0$(x+3)(x−3)=0.

Answer: $x = 3$x=3 or $x = -3$x=−3

Common error: writing the roots with the same sign as inside the bracket. From $(x + 2) = 0$(x+2)=0 the root is $x = -2$x=−2, not $+2$+2.

Worked Example 4

Solve $x^{2} = 6x$x2=6x.

Do not divide by $x$x (that would lose a root). Rearrange to $x^{2} - 6x = 0$x2−6x=0 and factorise: $x(x - 6) = 0$x(x−6)=0.

Answer: $x = 0$x=0 or $x = 6$x=6

Common error: cancelling the $x$x to get only $x = 6$x=6, throwing away the root $x = 0$x=0.

The Quadratic Formula

When a quadratic will not factorise neatly, use $x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$x=2a−b±b2−4ac​​. This formula solves any quadratic, whether or not it factorises, so it is your reliable fallback. Identify $a$a, $b$b and $c$c carefully, with their signs — writing them down explicitly before you substitute is a small habit that prevents many errors. The $\pm$± symbol is what produces the two roots: one using $+\sqrt{\ }$+ ​ and one using $-\sqrt{\ }$− ​.

Worked Example 5

Solve $x^{2} + 3x - 5 = 0$x2+3x−5=0, giving answers to $2$2 decimal places.

Here $a = 1$a=1, $b = 3$b=3, $c = -5$c=−5. Then $b^{2} - 4ac = 9 - 4(1)(-5) = 9 + 20 = 29$b2−4ac=9−4(1)(−5)=9+20=29.

$x = \dfrac{-3 \pm \sqrt{29}}{2}$x=2−3±29​​. Since $\sqrt{29} \approx 5.385$29​≈5.385, we get $x = \dfrac{-3 + 5.385}{2} \approx 1.19$x=2−3+5.385​≈1.19 or $x = \dfrac{-3 - 5.385}{2} \approx -4.19$x=2−3−5.385​≈−4.19.

Answer: $x \approx 1.19$x≈1.19 or $x \approx -4.19$x≈−4.19

Worked Example 6

Solve $2x^{2} - 5x - 1 = 0$2x2−5x−1=0, giving answers to $2$2 decimal places.

Here $a = 2$a=2, $b = -5$b=−5, $c = -1$c=−1. Then $b^{2} - 4ac = 25 - 4(2)(-1) = 25 + 8 = 33$b2−4ac=25−4(2)(−1)=25+8=33.

$x = \dfrac{5 \pm \sqrt{33}}{4}$x=45±33​​. Since $\sqrt{33} \approx 5.745$33​≈5.745, $x \approx \dfrac{5 + 5.745}{4} \approx 2.69$x≈45+5.745​≈2.69 or $x \approx \dfrac{5 - 5.745}{4} \approx -0.19$x≈45−5.745​≈−0.19.

Answer: $x \approx 2.69$x≈2.69 or $x \approx -0.19$x≈−0.19

Common error: mishandling the sign of $b$b. With $b = -5$b=−5, the formula uses $-b = +5$−b=+5 and $b^{2} = 25$b2=25.

The Discriminant

The discriminant $b^{2} - 4ac$b2−4ac tells you how many real roots a quadratic has: if it is positive there are two roots, if zero exactly one (a repeated root), and if negative there are no real roots.

Worked Example 7

How many real roots does $x^{2} + 4x + 7 = 0$x2+4x+7=0 have?

$b^{2} - 4ac = 16 - 4(1)(7) = 16 - 28 = -12$b2−4ac=16−4(1)(7)=16−28=−12, which is negative.

Answer: no real roots.

Completing the Square [Higher]

This whole section is Higher tier. To write $x^{2} + bx + c$x2+bx+c in the form $(x + p)^{2} + q$(x+p)2+q, take $p = \dfrac{b}{2}$p=2b​, then adjust the constant.

Worked Example 8

Write $x^{2} + 8x + 3$x2+8x+3 in the form $(x + p)^{2} + q$(x+p)2+q.

Half of $8$8 is $4$4, and $(x + 4)^{2} = x^{2} + 8x + 16$(x+4)2=x2+8x+16. We need only $+3$+3, so subtract $16$16 and add $3$3: $x^{2} + 8x + 3 = (x + 4)^{2} - 16 + 3 = (x + 4)^{2} - 13$x2+8x+3=(x+4)2−16+3=(x+4)2−13.

Answer: $(x + 4)^{2} - 13$(x+4)2−13

Worked Example 9

Solve $x^{2} + 8x + 3 = 0$x2+8x+3=0 by completing the square, giving exact answers.

From above, $(x + 4)^{2} - 13 = 0$(x+4)2−13=0, so $(x + 4)^{2} = 13$(x+4)2=13 and $x + 4 = \pm\sqrt{13}$x+4=±13​.

Answer: $x = -4 + \sqrt{13}$x=−4+13​ or $x = -4 - \sqrt{13}$x=−4−13​

Worked Example 10 — when $a \ne 1$a=1 [Higher]

Still Higher tier. Write $2x^{2} + 12x + 5$2x2+12x+5 in the form $a(x + p)^{2} + q$a(x+p)2+q.

Factor $2$2 from the $x$x-terms: $2[x^{2} + 6x] + 5$2[x2+6x]+5. Complete the square inside: $2[(x + 3)^{2} - 9] + 5 = 2(x + 3)^{2} - 18 + 5 = 2(x + 3)^{2} - 13$2[(x+3)2−9]+5=2(x+3)2−18+5=2(x+3)2−13.

Answer: $2(x + 3)^{2} - 13$2(x+3)2−13

Completing the square does more than solve equations: the form $(x + p)^{2} + q$(x+p)2+q instantly reveals the turning point of the parabola at $(-p, q)$(−p,q), which is why it is the method of choice when a question asks for a minimum or maximum value. It also produces exact roots in surd form, which the quadratic formula can give too but which factorising cannot when the roots are irrational.

Interpreting Roots in Context

In a real-world problem you must check that each root makes sense. A length or a time cannot be negative, so a negative root in such a context is rejected — but you must say so and give the reason, because the rejection itself earns a mark. This is where AO3 problem-solving meets AO2 communication.

Worked Example 11

A ball is thrown upward and its height $h$h metres after $t$t seconds is $h = 15t - 5t^{2}$h=15t−5t2. Work out when it returns to the ground.

Set $h = 0$h=0: $15t - 5t^{2} = 0$15t−5t2=0, so $5t(3 - t) = 0$5t(3−t)=0, giving $t = 0$t=0 or $t = 3$t=3. The start is $t = 0$t=0, so it lands at $t = 3$t=3.

Answer: after $3$3 seconds.

Worked Example 12

A rectangle has area $40\ \text{cm}^{2}$40 cm2. Its length is $3\ \text{cm}$3 cm more than its width. Work out the width.