Quadratic Equations Class 10: Notes and Important Questions

This guide covers Quadratic Equations Class 10, chapter 4 of NCERT Class 10 Mathematics. You will learn how to recognise a quadratic equation, solve it by factorisation and by the quadratic formula, and use the discriminant to predict the nature of its roots. Solved word problems, important questions and common mistakes are included.

What a quadratic equation is

A quadratic equation in x has the standard form

ax² + bx + c = 0, where a, b, c are real numbers and a ≠ 0.

Why a ≠ 0? If a = 0, the x² term disappears and the equation becomes bx + c = 0, which is at most linear.

A little history from the chapter: Brahmagupta gave an explicit formula for ax² + bx = c in the 7th century, and Sridharacharya derived the quadratic formula we use today around 1025 C.E.

Is it really quadratic? Simplify first

Always expand, collect like terms and bring everything to one side before deciding.

• x(x + 1) + 8 = (x + 2)(x − 2) becomes x² + x + 8 = x² − 4. The x² terms cancel, leaving x + 12 = 0. This is not quadratic.
• (x + 2)³ = x³ − 4 becomes x³ + 6x² + 12x + 8 = x³ − 4. The x³ terms cancel, leaving 6x² + 12x + 12 = 0. This is quadratic.
• (x − 2)² + 1 = 2x − 3 becomes x² − 4x + 5 = 2x − 3, so x² − 6x + 8 = 0. Quadratic, with a = 1, b = −6, c = 8.

Turning a situation into an equation

1. Let x stand for the unknown.
2. Write the other quantities in terms of x.
3. Use a relationship such as Area = Length × Breadth.
4. Simplify to ax² + bx + c = 0.

• A rectangular plot has area 528 m² and length one more than twice its breadth. Breadth = x, length = 2x + 1, so x(2x + 1) = 528, giving 2x² + x − 528 = 0.
• Two consecutive positive integers have product 306: x(x + 1) = 306, giving x² + x − 306 = 0.
• Two numbers have sum 27 and product 182: x(27 − x) = 182, giving x² − 27x + 182 = 0.

Roots and solving by factorisation

A real number α is a root of ax² + bx + c = 0 if aα² + bα + c = 0. Roots of p(x) = 0 are the same as the zeroes of the polynomial p(x). A quadratic equation has at most two roots.

Check: in 2x² − 3x + 1 = 0, putting x = 1 gives 2 − 3 + 1 = 0, so 1 is a root.

Splitting the middle term: find two numbers whose sum is b and whose product is ac. Factorise, then use the zero product property: if a product is zero, at least one factor is zero.

Example: 6x² − x − 2 = 0. We need product 6 × (−2) = −12 and sum −1: the numbers are −4 and 3.

• 6x² − 4x + 3x − 2 = 0
• 2x(3x − 2) + 1(3x − 2) = 0
• (2x + 1)(3x − 2) = 0
• x = −1/2 or x = 2/3

If both factors are the same, as in (√3x − √2)(√3x − √2) = 0, the root x = √(2/3) is repeated.

The quadratic formula

For ax² + bx + c = 0 with b² − 4ac ≥ 0:

x = (−b ± √(b² − 4ac)) / (2a)

1. Read off a, b and c.
2. Find D = b² − 4ac.
3. If D ≥ 0, use x = (−b ± √D) / (2a).
4. Work out the + and − cases separately.

Example: x² + 7x − 60 = 0. Here a = 1, b = 7, c = −60. D = 49 + 240 = 289 and √289 = 17. So x = (−7 + 17)/2 = 5 or x = (−7 − 17)/2 = −12.

Nature of roots: the discriminant

D = b² − 4ac is called the discriminant.

Example: for 2x² − 3x + 5 = 0, D = 9 − 40 = −31, so there are no real roots.

Word problems: check that answers make sense

Lengths, times and ages cannot be negative, so reject a negative root in such problems. If D < 0, the situation is impossible.

Pole in a park: a circular park has diameter 13 m. Gates A and B are at the ends of a diameter, and a pole P on the boundary is 7 m farther from A than from B. Angle APB is a right angle because AB is a diameter. Let BP = x, so AP = x + 7. Then x² + (x + 7)² = 13², which simplifies to x² + 7x − 60 = 0. The roots are 5 and −12. Reject −12. So BP = 5 m and AP = 5 + 7 = 12 m.

Ages: two friends' ages add up to 20, and 4 years ago the product of their ages was 48. With ages x and 20 − x: (x − 4)(16 − x) = 48 gives x² − 20x + 112 = 0. D = 400 − 448 = −48 < 0, so this situation is not possible.

Remember this

• Standard form: ax² + bx + c = 0 with a ≠ 0.
• Simplify fully before calling an equation quadratic.
• Splitting the middle term: two numbers with sum b and product ac.
• Formula: x = (−b ± √(b² − 4ac)) / (2a).
• D > 0 two distinct roots, D = 0 two equal roots, D < 0 no real roots.

Important questions with answers

1. Is x² + 3x + 1 = (x − 2)² a quadratic equation? No. The right side is x² − 4x + 4, so the x² terms cancel and leave 7x − 3 = 0, which is linear.

2. Solve x² − 3x − 10 = 0 by factorisation. Numbers with sum −3 and product −10 are −5 and 2. (x − 5)(x + 2) = 0, so x = 5 or x = −2.

3. Find the dimensions of the plot with area 528 m² and length one more than twice the breadth. 2x² + x − 528 = 0. D = 1 + 4(2)(528) = 1 + 4224 = 4225, √4225 = 65. x = (−1 + 65)/4 = 16 (the other root is negative). Breadth 16 m, length 2(16) + 1 = 33 m. Check: 16 × 33 = 528.

4. Find two consecutive positive integers whose product is 306. x² + x − 306 = 0. D = 1 + 1224 = 1225, √1225 = 35. x = (−1 + 35)/2 = 17. The integers are 17 and 18.

5. Two numbers have sum 27 and product 182. Find them. x² − 27x + 182 = 0. D = 729 − 728 = 1. x = (27 ± 1)/2 = 14 or 13. The numbers are 13 and 14.

6. Find the nature of the roots of x² − 6x + 9 = 0. D = 36 − 36 = 0, so there are two equal real roots, each −b/2a = 6/2 = 3.

7. Can a rectangular grove with length twice its breadth have area 800 m²? 2x² = 800, so x² = 400 and x = ±20. Breadth must be positive, so x = 20. Yes: breadth 20 m, length 40 m.

8. Why must a ≠ 0 in ax² + bx + c = 0? If a = 0, the x² term disappears and the equation is no longer quadratic.

Common mistakes to avoid

• Deciding the type of equation before simplifying.
• Forgetting the minus in −b, or the 2a in the denominator of the formula.
• Taking a real square root of a negative discriminant. D < 0 means no real roots.
• Keeping a negative root for a length, age or time.
• Using the wrong sign of c when finding D, for example writing 49 − 240 instead of 49 + 240 when c = −60.

Solve a few equations both by factorisation and by the formula to check yourself, and study this chapter with Joy whenever you need help.