Quadratic equations: Class 11 Mathematics Notes, Short Questions and Numericals

Standard form, solving by factoring, completing the square and the quadratic formula, the discriminant and nature of roots, sum and product of roots, forming equations, cube roots of unity and simultaneous equations.

Key points

A quadratic equation in x has the standard form ax² + bx + c = 0, where a, b and c are real numbers and a ≠ 0. A value of x that satisfies the equation is a root or solution. A quadratic equation has at most two roots.

Methods of solving ax² + bx + c = 0:

Factoring: split the middle term, write the left side as a product of two linear factors, and set each factor to 0.
Completing the square: divide by a, move the constant, add (b/2a)² to both sides, and take square roots.
Quadratic formula: x = (−b ± √(b² − 4ac)) / 2a.

The quadratic formula comes from completing the square. From x² + (b/a)x = −c/a, add (b/2a)² to both sides to get (x + b/2a)² = (b² − 4ac)/4a². Taking square roots gives the formula. It works for every quadratic, even when factoring is hard.

The expression D = b² − 4ac is the discriminant. It tells us the nature of the roots without solving. If D > 0 the roots are real and unequal. If D = 0 the roots are real and equal, each equal to −b/2a. If D < 0 the roots are not real, and they are a pair of complex conjugates. If a, b and c are rational and D is a perfect square, the roots are rational.

If α and β are the roots, then the sum is α + β = −b/a and the product is αβ = c/a. These follow by adding and multiplying the two roots from the formula. They let us find expressions in the roots without finding the roots. For example α² + β² = (α + β)² − 2αβ, and 1/α + 1/β = (α + β)/(αβ).

To form an equation with given roots α and β, use x² − (α + β)x + αβ = 0. If the sum is S and the product is P, the equation is x² − Sx + P = 0. Any multiple of this is also an equation with the same roots.

Some equations are not quadratic but can be reduced to quadratic form. A substitution such as y = x² turns x⁴ − 5x² + 4 = 0 into y² − 5y + 4 = 0, so y = 1 or 4 and x = ±1 or ±2. Equations with a radical, such as √(x + 2) = x, must be squared, and each root must be checked in the original equation because squaring can add extra roots.

The cube roots of unity are the roots of x³ = 1. Writing x³ − 1 = (x − 1)(x² + x + 1) = 0 gives x = 1, ω and ω², where ω = (−1 + i√3)/2 and ω² = (−1 − i√3)/2. Important properties are ω³ = 1, 1 + ω + ω² = 0, and ω² is the conjugate of ω. The cube roots of unity sum to 0 and have product 1.

Two equations in two unknowns, where one is quadratic, can be solved by substitution. Take the linear equation, solve for one variable, and substitute into the quadratic. This gives a quadratic in one variable, whose roots are then used to find the other variable. A pair of values (x, y) must satisfy both equations.

Word problems often lead to quadratic equations, for example on area, consecutive numbers, speed and time. Define the unknown, write the equation, solve it, and reject any root that does not fit the situation, such as a negative length.

Study tip: Write a, b and c on the first line of every solution and compute the discriminant before anything else. It tells you whether the roots will be real, and it avoids errors with signs. For sum and product questions, always rewrite the expression using (α + β) and αβ before substituting.

Definitions and flashcards

Say the meaning first, then open the card to check.

Quadratic equation

An equation of the form ax² + bx + c = 0 with a ≠ 0.

Root

A value of the variable that satisfies the equation.

Discriminant

The expression b² − 4ac, which decides the nature of the roots.

Complex conjugate roots

The pair p + qi and p − qi, which occur when D < 0 and the coefficients are real.

Radical equation

An equation in which the unknown is under a square root.

Cube roots of unity

The three numbers 1, ω and ω² whose cubes equal 1.

Extraneous root

A root that appears after squaring but fails in the original equation.

Formula sheet

Name	Formula	What it means
Quadratic formula	x = (−b ± √(b² − 4ac)) / 2a	Gives both roots.
Discriminant	D = b² − 4ac	Positive, zero or negative decides the nature.
Sum of roots	α + β = −b/a	From the coefficients.
Product of roots	αβ = c/a	From the coefficients.
Equation from roots	x² − (α + β)x + αβ = 0	Monic form.
Cube roots of unity	1 + ω + ω² = 0 and ω³ = 1	ω = (−1 + i√3)/2.

Short questions with model answers

These are the kind of questions asked for 2, 3 and 5 marks. Try each one before opening the answer.

Write the quadratic formula and state what the discriminant tells us. (2 marks)

For ax² + bx + c = 0, x = (−b ± √(b² − 4ac)) / 2a. The discriminant b² − 4ac tells the nature of the roots: positive for two real unequal roots, zero for two equal real roots, and negative for a pair of complex roots.

Find the sum and product of the roots of 3x² − 7x + 2 = 0. (2 marks)

Here a = 3, b = −7 and c = 2. The sum is −b/a = 7/3. The product is c/a = 2/3.

Find k so that x² + kx + 9 = 0 has equal roots. (3 marks)

Equal roots need D = 0. D = k² − 4(1)(9) = k² − 36. So k² = 36 and k = 6 or k = −6.

Why must the roots of a radical equation be checked? (3 marks)

Squaring both sides can create a new equation with more solutions than the original. For example, if x = −2 is squared we get x² = 4, which also allows x = 2. So each root found must be put back in the original equation, and any root that fails is rejected as extraneous.

Derive the quadratic formula by completing the square. (5 marks)

Start with ax² + bx + c = 0 and divide by a: x² + (b/a)x + c/a = 0. Move the constant: x² + (b/a)x = −c/a. Add (b/2a)² to both sides: (x + b/2a)² = b²/4a² − c/a = (b² − 4ac)/4a². Take square roots: x + b/2a = ±√(b² − 4ac) / 2a. Subtract b/2a from both sides to get x = (−b ± √(b² − 4ac)) / 2a.

State the properties of the cube roots of unity and prove that 1 + ω + ω² = 0. (5 marks)

The cube roots of unity are 1, ω and ω², with ω³ = 1, and ω² is the conjugate of ω. To prove the sum, note that x³ − 1 = (x − 1)(x² + x + 1). The roots ω and ω² are not 1, so they satisfy x² + x + 1 = 0. Putting x = ω gives ω² + ω + 1 = 0. So 1 + ω + ω² = 0.

Numericals with worked steps

Solve 2x² − 7x + 3 = 0 by factoring.

Find two numbers with product 2 × 3 = 6 and sum −7: they are −6 and −1.
Split the middle term: 2x² − 6x − x + 3 = 0.
Group: 2x(x − 3) − 1(x − 3) = 0.
Factor: (2x − 1)(x − 3) = 0.
So 2x − 1 = 0 or x − 3 = 0.

Answer: x = 1/2 or x = 3.

Solve x² − 4x − 1 = 0 using the quadratic formula.

Here a = 1, b = −4 and c = −1.
D = b² − 4ac = 16 + 4 = 20.
√20 = 2√5.
x = (4 ± 2√5) / 2 = 2 ± √5.

Answer: x = 2 + √5 or x = 2 − √5.

If α and β are the roots of 2x² − 5x + 1 = 0, find α² + β².

α + β = 5/2 and αβ = 1/2.
α² + β² = (α + β)² − 2αβ.
= 25/4 − 2(1/2) = 25/4 − 1.
= 25/4 − 4/4 = 21/4.

Answer: α² + β² = 21/4.

Form the quadratic equation whose roots are 2 + √3 and 2 − √3.

Sum = (2 + √3) + (2 − √3) = 4.
Product = (2)² − (√3)² = 4 − 3 = 1.
Equation: x² − (sum)x + product = 0.

Answer: x² − 4x + 1 = 0.

Solve x + y = 5 and xy = 6.

From the first equation, y = 5 − x.
Substitute: x(5 − x) = 6, so 5x − x² = 6.
Rearrange: x² − 5x + 6 = 0, so (x − 2)(x − 3) = 0.
If x = 2 then y = 3. If x = 3 then y = 2.

Answer: (x, y) = (2, 3) or (3, 2).

Common mistakes

Dividing both sides by x to cancel it, which loses the root x = 0.
Writing the quadratic formula with +b instead of −b, or dividing only part of the numerator by 2a.
Using the sum and product formulas before writing the equation in standard form with a ≠ 0.
Not checking the roots of a radical equation in the original equation.
Saying that D < 0 means there are no roots at all. There are no real roots, but there are two complex roots.

Practise this chapter

Practice questions and timed test Printable worksheet