Quadratic Equations • Topic 2 of 3

Solution by Factorization and Completing the Square

What does it mean to "solve" a quadratic equation? A real number \(\alpha\) is a root (or solution) of \(ax^{2}+bx+c=0\) if substituting \(x=\alpha\) makes the left side equal 0, i.e. \(a\alpha^{2}+b\alpha+c=0\). A quadratic has at most two roots. The roots are exactly the zeroes of the polynomial \(ax^{2}+bx+c\), which is why solving and factorising are two sides of the same coin.

1. Factorisation method (splitting the middle term). This rests on the zero-product rule: if a product of two factors is zero, then at least one of them must be zero. So once we write \(ax^{2}+bx+c=(px+q)(rx+s)=0\), we can set \(px+q=0\) and \(rx+s=0\) separately to get the two roots.

Steps for factorisation:

  • Write the equation in standard form \(ax^{2}+bx+c=0\).
  • Compute the product \(a \times c\).
  • Find two numbers whose product is \(a\times c\) and whose sum is \(b\).
  • Split the middle term \(bx\) using those two numbers.
  • Group in pairs and take out the common factor; a common binomial factor should appear.
  • Set each factor equal to zero and solve.

Finding the two numbers — a tip: match the sign of \(c\) (after using \(a\)). If \(a\times c\) is positive, the two numbers share the sign of \(b\); if \(a\times c\) is negative, they have opposite signs and the larger magnitude carries the sign of \(b\). For \(x^{2}-5x+6=0\), \(a\times c=6>0\) and \(b=-5<0\), so both numbers are negative: \(-2\) and \(-3\).

2. Completing the square. Not every quadratic factorises over nice integers, so we need a method that always works. Completing the square rewrites \(ax^{2}+bx+c=0\) as \((x+p)^{2}=q\), after which we simply take square roots. The key identity is \(x^{2}+2mx = (x+m)^{2}-m^{2}\): to "complete" \(x^{2}+bx\) into a perfect square we add \(\left(\dfrac{b}{2}\right)^{2}\).

Steps for completing the square:

  • If \(a\neq1\), divide the whole equation by \(a\) so the coefficient of \(x^{2}\) becomes 1.
  • Shift the constant term to the right side.
  • Add \(\left(\dfrac{b}{2}\right)^{2}\) to both sides.
  • Write the left side as a perfect square \((x+\tfrac{b}{2})^{2}\).
  • Take the square root of both sides, remembering the \(\pm\) sign.
  • Solve the two resulting linear equations.

3. The quadratic formula (Sridharacharya's rule). Completing the square on the general equation \(ax^{2}+bx+c=0\) gives a once-and-for-all formula. Divide by \(a\): \(x^{2}+\dfrac{b}{a}x+\dfrac{c}{a}=0\); add \(\left(\dfrac{b}{2a}\right)^{2}\) to both sides; the left becomes \(\left(x+\dfrac{b}{2a}\right)^{2}\) and the right becomes \(\dfrac{b^{2}-4ac}{4a^{2}}\). Taking square roots yields the celebrated formula:

\(x = \dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}\)

This works for every quadratic and is the safest method when factors are messy. The quantity \(b^{2}-4ac\) under the root (the discriminant) decides whether real roots exist — explored fully in the next topic.

Forming and solving word problems. After translating a word problem into standard form (Topic 1), you solve it here. Two cautions specific to word problems: always reject a root that is not physically meaningful (a negative length, a negative speed, a fractional number of people), and always state the answer in the units asked. Typical types: numbers and consecutive integers; rectangle area and dimensions; speed–distance–time with a time difference; combined work-rate and pipes filling a tank.

Common mistakes to avoid: cancelling a factor of \(x\) (which silently throws away the root \(x=0\)); forgetting the \(\pm\) after taking a square root; dividing by \(a\) but forgetting to divide the constant too; and accepting an invalid root in a word problem instead of rejecting it.

Solving a quadratic by factorisation using the zero product rule Method 1: factorisation (zero-product rule) x^2 + 7x + 12 = 0 product 12, sum 7 → 3 and 4 (x + 3)(x + 4) = 0 x + 3 = 0 x + 4 = 0 x = −3 x = −4 If a product of factors is 0, at least one factor must be 0. Completing the square shown geometrically by adding a small corner square Method 2: completing the square (x^2 + 6x) x^2 3x x 3 → x^2 3^2 x + 3 Add (b/2)^2 = 3^2 = 9 to fill the corner: x^2 + 6x + 9 = (x + 3)^2. The quadratic formula with the discriminant highlighted Method 3: the quadratic formula x = ( −b ± √(b^2 − 4ac) ) / 2a discriminant D = b^2 − 4ac Works for every quadratic; safest when factors are messy.
Solved Examples
1
Worked Example
Solve \(x^{2} - 5x + 6 = 0\) by factorisation.
Solution
  1. Step 1: Product \(a\times c = 1\times 6 = 6\); need two numbers with product 6 and sum \(-5\): they are \(-2\) and \(-3\).
  2. Step 2: Split the middle term: \(x^{2} - 2x - 3x + 6 = 0\).
  3. Step 3: Group: \(x(x - 2) - 3(x - 2) = 0\).
  4. Step 4: Factor out \((x - 2)\): \((x - 2)(x - 3) = 0\).
  5. Step 5: Zero-product rule: \(x - 2 = 0\) or \(x - 3 = 0\).

Answer: \(x = 2,\ x = 3\)

2
Worked Example
Solve \(6x^{2} - x - 2 = 0\) by factorisation.
Solution
  1. Step 1: \(a\times c = 6\times(-2) = -12\); need two numbers with product \(-12\) and sum \(-1\): they are \(-4\) and \(3\).
  2. Step 2: Split: \(6x^{2} - 4x + 3x - 2 = 0\).
  3. Step 3: Group: \(2x(3x - 2) + 1(3x - 2) = 0\).
  4. Step 4: Factor: \((3x - 2)(2x + 1) = 0\).
  5. Step 5: \(3x - 2 = 0 \Rightarrow x = \dfrac{2}{3}\); or \(2x + 1 = 0 \Rightarrow x = -\dfrac{1}{2}\).

Answer: \(x = \dfrac{2}{3},\ x = -\dfrac{1}{2}\)

3
Worked Example
Solve \(2x^{2} + x - 6 = 0\) by factorisation.
Solution
  1. Step 1: \(a\times c = 2\times(-6) = -12\); two numbers with product \(-12\) and sum \(1\): they are \(4\) and \(-3\).
  2. Step 2: Split: \(2x^{2} + 4x - 3x - 6 = 0\).
  3. Step 3: Group: \(2x(x + 2) - 3(x + 2) = 0\).
  4. Step 4: Factor: \((x + 2)(2x - 3) = 0\).
  5. Step 5: \(x + 2 = 0 \Rightarrow x = -2\); or \(2x - 3 = 0 \Rightarrow x = \dfrac{3}{2}\).

Answer: \(x = -2,\ x = \dfrac{3}{2}\)

4
Worked Example
Solve \(\sqrt{2}\,x^{2} + 7x + 5\sqrt{2} = 0\) by factorisation.
Solution
  1. Step 1: \(a\times c = \sqrt{2}\times 5\sqrt{2} = 5\times 2 = 10\); need product 10, sum 7: they are 5 and 2.
  2. Step 2: Split: \(\sqrt{2}\,x^{2} + 5x + 2x + 5\sqrt{2} = 0\).
  3. Step 3: Group: \(x(\sqrt{2}\,x + 5) + \sqrt{2}(\sqrt{2}\,x + 5) = 0\) (using \(2 = \sqrt{2}\cdot\sqrt{2}\)).
  4. Step 4: Factor: \((\sqrt{2}\,x + 5)(x + \sqrt{2}) = 0\).
  5. Step 5: \(\sqrt{2}\,x + 5 = 0 \Rightarrow x = -\dfrac{5}{\sqrt{2}}\); or \(x + \sqrt{2} = 0 \Rightarrow x = -\sqrt{2}\).

Answer: \(x = -\dfrac{5}{\sqrt{2}},\ x = -\sqrt{2}\)

5
Worked Example
Solve \(x^{2} - 4x - 12 = 0\) by completing the square.
Solution
  1. Step 1: Move the constant to the right: \(x^{2} - 4x = 12\).
  2. Step 2: Here \(b = -4\), so \(\left(\dfrac{b}{2}\right)^{2} = (-2)^{2} = 4\).
  3. Step 3: Add 4 to both sides: \(x^{2} - 4x + 4 = 12 + 4\).
  4. Step 4: Left side is a perfect square: \((x - 2)^{2} = 16\).
  5. Step 5: Take square roots: \(x - 2 = \pm 4\).
  6. Step 6: \(x = 2 + 4 = 6\) or \(x = 2 - 4 = -2\).

Answer: \(x = 6,\ x = -2\)

6
Worked Example
Solve \(2x^{2} - 7x + 3 = 0\) by completing the square.
Solution
  1. Step 1: Divide by 2: \(x^{2} - \dfrac{7}{2}x + \dfrac{3}{2} = 0\).
  2. Step 2: Move constant: \(x^{2} - \dfrac{7}{2}x = -\dfrac{3}{2}\).
  3. Step 3: Half of \(-\dfrac{7}{2}\) is \(-\dfrac{7}{4}\); its square is \(\dfrac{49}{16}\).
  4. Step 4: Add to both sides: \(x^{2} - \dfrac{7}{2}x + \dfrac{49}{16} = -\dfrac{3}{2} + \dfrac{49}{16} = \dfrac{-24 + 49}{16} = \dfrac{25}{16}\).
  5. Step 5: \(\left(x - \dfrac{7}{4}\right)^{2} = \dfrac{25}{16}\Rightarrow x - \dfrac{7}{4} = \pm\dfrac{5}{4}\).
  6. Step 6: \(x = \dfrac{7}{4} + \dfrac{5}{4} = 3\) or \(x = \dfrac{7}{4} - \dfrac{5}{4} = \dfrac{1}{2}\).

Answer: \(x = 3,\ x = \dfrac{1}{2}\)

7
Worked Example
Solve \(x^{2} + 4x - 5 = 0\) using the quadratic formula.
Solution
  1. Step 1: Identify \(a = 1,\ b = 4,\ c = -5\).
  2. Step 2: Discriminant \(b^{2} - 4ac = 16 - 4(1)(-5) = 16 + 20 = 36\).
  3. Step 3: \(\sqrt{36} = 6\).
  4. Step 4: \(x = \dfrac{-b \pm \sqrt{36}}{2a} = \dfrac{-4 \pm 6}{2}\).
  5. Step 5: \(x = \dfrac{2}{2} = 1\) or \(x = \dfrac{-10}{2} = -5\).

Answer: \(x = 1,\ x = -5\)

8
Worked Example
Solve \(3x^{2} - 5x + 2 = 0\) using the quadratic formula.
Solution
  1. Step 1: \(a = 3,\ b = -5,\ c = 2\).
  2. Step 2: \(b^{2} - 4ac = 25 - 4(3)(2) = 25 - 24 = 1\).
  3. Step 3: \(\sqrt{1} = 1\).
  4. Step 4: \(x = \dfrac{5 \pm 1}{6}\).
  5. Step 5: \(x = \dfrac{6}{6} = 1\) or \(x = \dfrac{4}{6} = \dfrac{2}{3}\).

Answer: \(x = 1,\ x = \dfrac{2}{3}\)

9
Worked Example
The product of two consecutive positive integers is 156. Find the integers.
Solution
  1. Step 1: Let the smaller integer be \(x\); the next is \(x + 1\), so \(x(x + 1) = 156\).
  2. Step 2: Standard form: \(x^{2} + x - 156 = 0\).
  3. Step 3: Factorise: need product \(-156\), sum 1: \(13\) and \(-12\). So \(x^{2} + 13x - 12x - 156 = 0\).
  4. Step 4: Group: \(x(x + 13) - 12(x + 13) = 0 \Rightarrow (x + 13)(x - 12) = 0\).
  5. Step 5: \(x = -13\) or \(x = 12\). Reject \(-13\) (must be positive), so \(x = 12\).
  6. Step 6: The integers are 12 and 13.

Answer: 12 and 13

10
Worked Example
The area of a rectangular plot is 528 m². The length is one more than twice the breadth. Find the length and breadth.
Solution
  1. Step 1: Let the breadth be \(x\) m; then the length is \((2x + 1)\) m.
  2. Step 2: Area: \(x(2x + 1) = 528\Rightarrow 2x^{2} + x - 528 = 0\).
  3. Step 3: \(a\times c = -1056\); two numbers with product \(-1056\), sum 1: \(33\) and \(-32\).
  4. Step 4: Split and group: \(2x^{2} + 33x - 32x - 528 = 0 \Rightarrow x(2x + 33) - 16(2x + 33) = 0\).
  5. Step 5: \((2x + 33)(x - 16) = 0 \Rightarrow x = -\dfrac{33}{2}\) or \(x = 16\). Reject the negative breadth.
  6. Step 6: Breadth \(= 16\) m; length \(= 2(16) + 1 = 33\) m.

Answer: Breadth = 16 m, Length = 33 m

11
Worked Example
A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less. Find the speed of the train.
Solution
  1. Step 1: Let the speed be \(x\) km/h. Time \(= \dfrac{360}{x}\); at speed \(x + 5\), time \(= \dfrac{360}{x + 5}\), which is 1 hour less.
  2. Step 2: \(\dfrac{360}{x} - \dfrac{360}{x + 5} = 1\).
  3. Step 3: Multiply by \(x(x + 5)\): \(360(x + 5) - 360x = x(x + 5)\), i.e. \(1800 = x^{2} + 5x\).
  4. Step 4: Standard form: \(x^{2} + 5x - 1800 = 0\).
  5. Step 5: Factorise (product \(-1800\), sum 5: \(45, -40\)): \((x + 45)(x - 40) = 0\).
  6. Step 6: \(x = -45\) (rejected) or \(x = 40\).

Answer: Speed of the train = 40 km/h

12
Worked Example
Two water taps together can fill a tank in \(9\dfrac{3}{8}\) hours. The larger tap takes 10 hours less than the smaller tap to fill the tank alone. Find the time each tap takes alone.
Solution
  1. Step 1: Let the smaller tap take \(x\) hours alone; the larger takes \((x - 10)\) hours.
  2. Step 2: In 1 hour they fill \(\dfrac{1}{x}\) and \(\dfrac{1}{x - 10}\) of the tank; together \(\dfrac{1}{x} + \dfrac{1}{x - 10}\). The combined time is \(9\dfrac{3}{8} = \dfrac{75}{8}\) hours, so combined rate \(= \dfrac{8}{75}\).
  3. Step 3: \(\dfrac{1}{x} + \dfrac{1}{x - 10} = \dfrac{8}{75}\), giving \(\dfrac{(x - 10) + x}{x(x - 10)} = \dfrac{8}{75}\), i.e. \(\dfrac{2x - 10}{x^{2} - 10x} = \dfrac{8}{75}\).
  4. Step 4: Cross-multiply: \(75(2x - 10) = 8(x^{2} - 10x)\Rightarrow 150x - 750 = 8x^{2} - 80x\).
  5. Step 5: Standard form: \(8x^{2} - 230x + 750 = 0\); divide by 2: \(4x^{2} - 115x + 375 = 0\).
  6. Step 6: Quadratic formula: \(b^{2} - 4ac = 13225 - 6000 = 7225\), \(\sqrt{7225} = 85\); \(x = \dfrac{115 \pm 85}{8}\).
  7. Step 7: \(x = \dfrac{200}{8} = 25\) or \(x = \dfrac{30}{8} = 3.75\). Reject 3.75 (then \(x - 10\) is negative), so \(x = 25\).
  8. Step 8: Smaller tap = 25 hours; larger tap = \(25 - 10 = 15\) hours.

Answer: Smaller tap takes 25 hours, larger tap takes 15 hours

Key Points

  • A root of \(ax^{2}+bx+c=0\) is a value of \(x\) that makes the expression 0; a quadratic has at most two roots
  • Factorisation uses the zero-product rule: if \((px+q)(rx+s)=0\) then \(px+q=0\) or \(rx+s=0\)
  • Splitting the middle term: find two numbers with product \(a\times c\) and sum \(b\)
  • Completing the square works for all quadratics: add \(\left(\dfrac{b}{2}\right)^{2}\) after making the \(x^{2}\) coefficient 1
  • Quadratic formula (Sridharacharya): \(x = \dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}\)
  • In word problems, always reject roots that are physically impossible (negative length, speed, or count)
  • Never cancel a common factor of \(x\) — that loses the root \(x=0\); always check by substitution
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.The roots of $x^2-5x+6=0$ are:
Explanation: $(x-2)(x-3)=0$.
Q2.The roots of $x^2-4=0$ are:
Explanation: $x^2=4$.
Q3.To complete the square of $x^2+6x$, add:
Explanation: $(6/2)^2=9$.
Q4.The roots of $(x-1)(x-4)=0$ are:
Explanation: Set factors to zero.
Practice more MCQs → 📝 Assignment · 35 marks