Quantitative Aptitude · Equations

Sum and Product of Roots of a Quadratic Equation (CA Foundation)

Updated 1 October 2026 · Fact-checked

For ax² + bx + c = 0 with roots α and β, the sum is α + β = −b/a and the product is αβ = c/a. You read these from the coefficients, without solving. Then rewrite any asked expression using only α + β and αβ, or build a new equation using x² − (sum)x + product = 0.

Understand Sum and Product of Roots

A quadratic equation ax² + bx + c = 0 has two roots, called α and β. You can find them by factorising or by the formula. But many exam questions never ask for the roots themselves. They ask for something built from them, such as α² + β² or 1/α + 1/β.

The key idea is that the roots are tied to the coefficients. If the roots are α and β, the equation can be written as a(x − α)(x − β) = 0. Expand it: a[x² − (α + β)x + αβ] = 0. Compare this with ax² + bx + c = 0. You get α + β = −b/a and αβ = c/a.

So the sum and product are known the moment you see the equation. Everything else is algebra. Any symmetric expression (one that stays the same if you swap α and β) can be rewritten using just the sum and the product. For example, α² + β² = (α + β)² − 2αβ.

The same idea works backwards. If you are given two roots, or any facts about them, you find the sum and product and write x² − (sum)x + product = 0. To form an equation whose roots are related to α and β (like 2α and 2β, or 1/α and 1/β), find the new sum and new product from the old ones.

Key formulas to remember

Sum of roots: α + β = −b/a; For ax² + bx + c = 0 with a ≠ 0. Watch the minus sign.
Product of roots: αβ = c/a; No minus sign. Divide by a, the coefficient of x².
Equation from roots: x² − (α + β)x + αβ = 0; Multiply by a common denominator if the coefficients are fractions.
Sum of squares: α² + β² = (α + β)² − 2αβ; Most common expression in exams.
Difference of roots: (α − β)² = (α + β)² − 4αβ; Also (α − β)² = D/a², where D = b² − 4ac. Take the square root carefully for the sign.
Sum of cubes: α³ + β³ = (α + β)³ − 3αβ(α + β); Also written (α + β)(α² − αβ + β²).
Sum of reciprocals: 1/α + 1/β = (α + β) ÷ αβ; Valid only when αβ ≠ 0, that is c ≠ 0.
Ratio form: α/β + β/α = (α² + β²) ÷ αβ; Find α² + β² first.
Equation with reciprocal roots: cx² + bx + a = 0; Swap a and c. Needs c ≠ 0.
Equation with negative roots: ax² − bx + c = 0; Roots are −α and −β. Only the sign of b changes.
Equation with roots kα and kβ: ax² + kbx + k²c = 0; Sum becomes k times the old sum, product k² times the old product.

How to solve Sum and Product of Roots questions

Use this method for any question on roots and coefficients. Do not solve for the roots unless the question forces you to.

1Write the equation in standard form ax² + bx + c = 0. Move all terms to one side and identify a, b and c with their signs.
2Find the sum S = −b/a and the product P = c/a.
3Look at the expression asked. Rewrite it using only S and P with an identity such as α² + β² = S² − 2P.
4Substitute S and P and calculate. Keep fractions exact until the end.
5If you must form a new equation, find the new sum and new product from S and P, then write x² − (new sum)x + new product = 0.
6Clear fractions by multiplying through, so the coefficients are integers.
7Check against the options. A quick test: the sum of roots of your new equation should match what you worked out.

Quickest way: Sum-product shortcut with option elimination

When to use it: Use in the MCQ paper whenever the question asks for an equation or a value from the roots. It takes under a minute.

Get S = −b/a and P = c/a in one line.
For a value, plug into the identity directly. Memorise α² + β² = S² − 2P and α³ + β³ = S³ − 3PS.
For a new equation, compute the new sum and product, then test each option. For options x² + px + q = 0 (a = 1), p must equal −(new sum) and q must equal new product.
Check signs first. Most wrong options differ only in the sign of the middle term, so one sign check often removes two options.
If the equation has a ≠ 1, make sure you divided by a. Options that look like the original coefficients are usually traps.
If the algebra takes more than about two minutes, skip it and return later. A wrong answer costs 0.25 marks.

Common mistakes in Sum and Product of Roots

Taking the sum of roots as b/a instead of −b/a.
The minus sign is easy to forget when you rush, and the product has no minus sign, so the two rules feel alike.
Fix: Say it as "sum is minus b over a, product is c over a". Test with x² − 5x + 6 = 0: roots 2 and 3, sum 5 = −(−5)/1.
Forgetting to divide by a when a ≠ 1.
Students see 2x² − 6x + 3 = 0 and use −b = 6 as the sum.
Fix: Always write S = −b/a and P = c/a with the actual value of a. Here S = 3 and P = 3/2.
Writing the new equation as x² − Sx − P = 0 or x² + Sx + P = 0.
Mixing up the signs of the standard form.
Fix: The form is always x² − (sum)x + (product) = 0. Check by expanding (x − α)(x − β).
Using α² + β² = (α + β)² and skipping the −2αβ.
It feels natural to square the sum and stop.
Fix: Remember (α + β)² = α² + 2αβ + β². So you must subtract 2αβ to leave α² + β².
Mixing up the sum and product when finding the equation with reciprocal roots.
Students invert the whole equation or swap the wrong things.
Fix: New sum = S ÷ P and new product = 1 ÷ P. Compute both, then write the equation.
Leaving fractional coefficients and picking the wrong option.
Options are given with integer coefficients, but your equation has fractions.
Fix: Multiply through by the common denominator. x² − (5/2)x + 1/2 = 0 becomes 2x² − 5x + 1 = 0.

Worked examples

Example 1

If α and β are the roots of 2x² − 6x + 3 = 0, the value of α² + β² is: (a) 3 (b) 6 (c) 9 (d) 12

Show the solution

Here a = 2, b = −6, c = 3.
Sum S = −b/a = 6/2 = 3.
Product P = c/a = 3/2.
α² + β² = S² − 2P = 3² − 2 × (3/2) = 9 − 3 = 6.

Answer: (b) 6

Example 2

If α and β are the roots of x² − 5x + 2 = 0, the equation whose roots are 1/α and 1/β is: (a) x² − 5x + 2 = 0 (b) 2x² − 5x + 1 = 0 (c) 2x² + 5x + 1 = 0 (d) x² − 2x + 5 = 0

Show the solution

For the given equation, S = 5 and P = 2.
New sum = 1/α + 1/β = S ÷ P = 5/2.
New product = (1/α)(1/β) = 1/P = 1/2.
The equation is x² − (5/2)x + 1/2 = 0.
Multiply by 2: 2x² − 5x + 1 = 0.
Check with the swap rule: swapping a and c in x² − 5x + 2 = 0 gives 2x² − 5x + 1 = 0. It matches.

Answer: (b) 2x² − 5x + 1 = 0

Example 3

If α and β are the roots of x² − 6x + 4 = 0, the value of α³ + β³ is: (a) 72 (b) 144 (c) 180 (d) 216

Show the solution

Here S = −(−6)/1 = 6 and P = 4/1 = 4.
Use α³ + β³ = S³ − 3PS.
S³ = 216.
3PS = 3 × 4 × 6 = 72.
α³ + β³ = 216 − 72 = 144.

Answer: (b) 144

Exam tips

Questions on this topic are usually one-step or two-step identity problems. Memorise α² + β² and α³ + β³ so you do not derive them in the exam.
When asked for the equation, check the sign of the middle term and the constant term in the options before doing anything else. This often cuts the options to two.
Do not solve for the roots unless the roots are clearly whole numbers. Using S and P is faster and avoids square-root errors.
If a problem gives a condition like "sum of roots is 4" and asks for a constant such as k, set −b/a equal to 4 and solve for k. Then verify that the other conditions still hold.
Keep fractions exact. Convert to decimals only if the options are decimals.

Practice questions from Equations

Sum and Product of Roots: frequently asked questions

What is the formula for the sum and product of roots of a quadratic equation?

For ax² + bx + c = 0, the sum of roots is −b/a and the product is c/a. This holds when a ≠ 0. The formulas work whether the roots are real or not.

How do I form a quadratic equation from given roots?

Find the sum and product of the given roots. Then write x² − (sum)x + product = 0. If you get fractions, multiply through to make the coefficients whole numbers.

How do I find α² + β² without solving the equation?

Find S = α + β and P = αβ from the coefficients. Then use α² + β² = S² − 2P. The same method works for the cubes, reciprocals and differences of roots.

Does this work if the roots are not real?

Yes. The relations between roots and coefficients hold for all roots of the equation, real or complex. In CA Foundation, questions mostly use real roots, but you do not need to check the discriminant to use these formulas.