Quadratic equations for CDS

Quadratic questions in CDS are rarely "solve this equation" alone. More often they ask about the roots without asking you to find them: the sum of their squares, the equation whose roots are their squares, the value of k that makes them equal. All of these come from two facts, the sum and the product of the roots, plus the discriminant.

Learn those three quantities well and most quadratic questions take under a minute.

Solving a quadratic

For ax² + bx + c = 0 with a ≠ 0:

• Factorisation: split the middle term using two numbers whose product is ac and whose sum is b.
• Formula: x = (−b ± √D) ÷ (2a), where D = b² − 4ac.

Factorisation is quicker when the numbers are friendly. The formula always works. Use it when you cannot find the split in a few seconds.

Sum and product of roots

If α and β are the roots of ax² + bx + c = 0:

• α + β = −b/a
• αβ = c/a

Why: the equation can be written as a(x − α)(x − β) = 0. Expanding gives ax² − a(α + β)x + aαβ. Matching coefficients with ax² + bx + c gives both results.

It follows that the quadratic with roots α and β is x² − (sum)x + (product) = 0.

Expressions in α and β

Any symmetric expression can be written using the sum and product alone.

These are the algebraic identities you already know, applied to roots. See the identities guide for why they hold.

The discriminant and nature of roots

Some quick facts about roots, useful for option elimination:

• Roots are reciprocals of each other when c = a (the product is 1).
• Roots are equal in size and opposite in sign when b = 0 (the sum is 0).
• One root is zero when c = 0.
• If a and c have opposite signs, D is automatically positive and the roots have opposite signs.

Infinite nested roots

An expression such as √(6 + √(6 + √(6 + …))) repeats itself inside. Call it x. Then x = √(6 + x), so x² − x − 6 = 0, and (x − 3)(x + 2) = 0. The value must be positive, so x = 3.

Worked questions

Question 1: Solve 2x² − 7x + 3 = 0.

• D = 49 − 24 = 25, so √D = 5.
• x = (7 ± 5) ÷ 4, so the roots are 3 and 1/2.
• By factorisation: ac = 6 and 6 + 1 = 7, so 2x² − 6x − x + 3 = (2x − 1)(x − 3).

Question 2: α and β are the roots of x² − 6x + 4 = 0. Find α² + β², 1/α + 1/β and (α − β)².

• S = 6 and P = 4.
• α² + β² = 36 − 8 = 28.
• 1/α + 1/β = 6/4 = 3/2.
• (α − β)² = 36 − 16 = 20.

Question 3: Form the equation whose roots are the squares of the roots of x² − 3x + 1 = 0.

• S = 3 and P = 1 for the given roots.
• New sum = α² + β² = 9 − 2 = 7. New product = α²β² = 1.
• The equation is x² − 7x + 1 = 0.

Question 4: For what values of k does x² + kx + 16 = 0 have equal roots?

• D = k² − 64 = 0, so k² = 64.
• k = 8 or k = −8.

Question 5: One root of x² − 9x + c = 0 is twice the other. Find c.

• Let the roots be r and 2r. Their sum 3r = 9, so r = 3 and the roots are 3 and 6.
• c = product = 18.

Question 6: The sum of a number and its reciprocal is 10/3. Find the number.

• x + 1/x = 10/3, so 3x² − 10x + 3 = 0.
• (3x − 1)(x − 3) = 0, so the number is 3 or 1/3.

Practice set

1. Find the roots of x² − x − 12 = 0.
2. Form the quadratic equation whose roots are 3 and −5.
3. What is the nature of the roots of 2x² + 3x + 5 = 0?
4. One root of x² − 5x + k = 0 is 2. Find k and the other root.
5. α and β are the roots of x² − 4x + 2 = 0. Find α² + β².
6. For what value of k are the roots of 3x² − 6x + k = 0 equal?
7. Find the value of √(12 + √(12 + √(12 + …))).
8. The roots of x² + px + 12 = 0 differ by 1. Find p.

Answers

1. 4 and −3. (x − 4)(x + 3) = 0.
2. x² + 2x − 15 = 0. Sum = −2 and product = −15.
3. Not real. D = 9 − 40 = −31.
4. k = 6, other root 3. 4 − 10 + k = 0. The sum of the roots is 5.
5. 12. 16 − 2 × 2.
6. k = 3. 36 − 12k = 0.
7. 4. x² = 12 + x gives (x − 4)(x + 3) = 0, and x must be positive.
8. p = 7 or p = −7. (α − β)² = p² − 48 = 1, so p² = 49. The roots are −3 and −4, or 3 and 4.

What to do next

• Learn the S and P table and derive each line once yourself
• Practise ten "without finding the roots" questions
• For every k question, check whether a ± answer is possible
• Revise linear equations and then move to logarithms and indices