Learning Objectives
Choose factorisation, completing the square or the quadratic formula.
Convert a quadratic to a constant plus a squared expression.
Find exact real roots and recognise when none exist.
Check both roots and restrictions in the original problem.
Prerequisites: Section 1B factorisation and equations; Section 1A surds. All practice questions in this lesson are original, not official TMUA questions.
What This Lesson Is About
A quadratic equation contains a nonzero squared term and asks which inputs make the expression zero. This lesson develops three ways to find those inputs: factorisation, completing the square and the quadratic formula.
Completing the square also explains why an expression has a minimum or maximum and why some equations have no real solutions. We finish by checking which algebraic roots are allowed in a practical problem.
1. Recognising and Factoring a Quadratic Equation
Put the equation into a standard form before choosing a method. A visible factorisation gives a direct route to the roots.
1.1 Identifying the coefficients in standard form
A quadratic equation can be written ax² + bx + c = 0 with a ≠ 0. The coefficients a, b and c are constants with respect to x. Rearrange to zero before identifying them; a negative coefficient includes its sign.
1.2 Finding roots with the zero-product rule
If a factorisation is visible, use the zero-product rule. For example, x² − 5x + 6 = (x − 2)(x − 3), so the roots are 2 and 3. Dividing by either factor would lose the root making it zero.
Worked Example: A Non-Monic Quadratic
Question: Solve 2x² + x − 3 = 0.
Method: Factor the quadratic into two linear factors and set each factor equal to zero.
Factor as (2x + 3)(x − 1) = 0. Therefore x = −3/2 or x = 1.
Check: at −3/2 the expression is 9/2 − 3/2 − 3 = 0; at 1 it is 2 + 1 − 3 = 0.
2. Completing the Square
When factorisation is less obvious, rewrite the expression using a single squared bracket. Balancing the constant keeps the rewritten expression equal to the original.
2.1 Choosing half the linear coefficient
Expansion gives (x + h)² = x² + 2hx + h². To represent x² + bx as a square, choose h = b/2 and subtract the extra constant: x² + bx = (x + b/2)² − b²/4.
2.2 Keeping the outer coefficient and interpreting the square
When a ≠ 1, first factor a out of the quadratic and linear terms. Keep track of the multiplier on the added and subtracted constant. This representation also shows a minimum or maximum because a real square is nonnegative.
Worked Example: Keep the Outer Multiplier
Question: Rewrite 2x² − 12x + 7 and solve 2x² − 12x + 7 = 0.
Method: Factor 2 from the quadratic and linear terms, complete the square inside the bracket, then solve by taking both square-root signs.
2(x² − 6x) + 7 = 2[(x − 3)² − 9] + 7 = 2(x − 3)² − 11.
The equation becomes (x − 3)² = 11/2, so x = 3 ± √22/2. Both signs are needed because either displacement has the same square.
Substitution into 2(x − 3)² − 11 gives 2(22/4) − 11 = 0. The minimum value of the expression is −11 at x = 3.
3. Understanding the Quadratic Formula
Applying the same square-completion steps to general coefficients produces a formula that works for every genuine quadratic. The quantity under its square root determines whether real answers exist.
3.1 Deriving the formula by completing the square
Divide ax² + bx + c = 0 by nonzero a, move c/a to the other side and complete the square. This gives (x + b/(2a))² = (b² − 4ac)/(4a²). Equivalently, (2ax + b)² = b² − 4ac.
3.2 Substituting coefficients and checking for real roots
Taking both square-root signs gives x = (−b ± √(b² − 4ac))/(2a), provided b² − 4ac ≥ 0 for real roots. Parenthesise the entire numerator and denominator. If the expression under the root is negative, there are no real solutions.
3.3 Interpreting the two square-root signs
The symbol ± abbreviates two candidates, one with addition and one with subtraction. If the square root is zero, they coincide and there is one distinct root.
Worked Example: Exact Irrational Roots
Question: Solve 3x² + 2x − 2 = 0.
Method: Identify a, b and c with their signs, substitute into the formula, and simplify the surd exactly.
Here a = 3, b = 2 and c = −2. The square-root expression is 4 − 4(3)(−2) = 28.
x = (−2 ± √28)/6 = (−1 ± √7)/3. Check independently by completing the square: (3x + 1)² = 7. Both displayed roots satisfy this.
Worked Example: No Real Root
Question: Solve x² + 4x + 8 = 0 over the reals.
Method: Complete the square to find the smallest possible value of the left side and decide whether it can reach zero.
The left side is (x + 2)² + 4, which is at least 4. It cannot be zero.
The formula agrees: b² − 4ac = 16 − 32 = −16, a negative value.
4. Checking Roots Against the Context
Solving the equation produces algebraic candidates. A model or restricted expression can impose extra conditions that remove some of them.
4.1 Applying the original restrictions
A quadratic model may give two algebraic roots but only one physically allowed value. Write the restrictions before solving and check them afterwards. A length must be positive; a rational expression must retain its original denominator exclusions.
Worked Example: A Rectangle
Question: A rectangle has width w metres and length w + 1 metres, with area 12 m². Find its dimensions.
Method: Write area as width times length, solve the resulting quadratic, and keep only a positive width.
w(w + 1) = 12 gives w² + w − 12 = (w + 4)(w − 3) = 0.
The roots are −4 and 3, but w > 0. Thus width is 3 m and length is 4 m; their product is 12 m².
Common Mistakes
Do not forget to multiply the square-completion adjustment by the leading coefficient.
A square equation with a positive right side has two roots, not just the principal root.
Keep the entire 2a in the formula denominator.
State whether the question asks for real roots and apply any contextual restrictions.
5. Practice
Use the form of each expression to choose factorisation, completing the square or the formula. Give exact roots, include both square-root signs when needed, and apply any restrictions stated in the question.
1. Solve x² − 7x + 10 = 0.
2. Write x² + 6x + 11 in completed-square form and state its minimum.
3. Solve 2x² − 4x − 1 = 0 exactly.
4. Solve 4x² + 4x + 1 = 0.
5. Explain why 3x² − 6x + 5 = 0 has no real root.
6. A rectangle has width w > 0 and length w + 3, with area 10. Find both dimensions.
Worked Solutions
1. Solving by factorisation
(x − 2)(x − 5) = 0, so x = 2 or 5. Substitution gives 4 − 14 + 10 = 0 and 25 − 35 + 10 = 0.
2. Reading a minimum from a completed square
(x + 3)² + 2 has minimum 2 at x = −3, since its squared term is nonnegative and can be zero.
3. Finding two exact irrational roots
2(x − 1)² − 3 = 0 gives x = 1 ± √6/2. Check: each displacement square is 6/4, giving 2(6/4) − 3 = 0.
4. Recognising a repeated root
(2x + 1)² = 0 gives the one distinct root x = −1/2. The original equals 1 − 2 + 1 = 0.
5. Proving there is no real root
3x² − 6x + 5 = 3(x − 1)² + 2 ≥ 2, so it cannot vanish.
6. Selecting the positive dimensions
w² + 3w − 10 = (w + 5)(w − 2) = 0. Only w = 2 is positive; the length is 5. The product is 10.
6. Summary
- Use factorisation when visible and complete a square to expose structure.
- The quadratic formula supplies exact roots when the leading coefficient is nonzero.
- A nonnegative square distinguishes two, one or no real roots.
- Check each candidate against the equation and its domain.