NCERT Class 10 Maths Exercise 2.2 Solution – Polynomials

Open the questions in the official book: NCERT Class 10 Maths Chapter 2 (PDF) or read it on the NCERT website.

Video Solution – Exercise 2.2

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Step-by-Step Solutions – Exercise 2.2

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Question 1

Question 1 (i)

Find the zeroes of $x^2-2x-8$ and verify the relationship between the zeroes and the coefficients.

Solution:

1. $$x^2-2x-8=x^2-4x+2x-8$$

   Split the middle term. Write -2x as -4x + 2x, because their product is -8x².

2. $$=x(x-4)+2(x-4)=(x-4)(x+2)$$

   Take out common factors to get (x – 4)(x + 2).

3. $$\text{Zeroes: }4,-2$$

   Setting these to zero gives x = 4 and x = -2. These are the two zeroes.

4. $$\text{Sum}=4+(-2)=2=-{\displaystyle \frac{-2}{1}}\mathrm{✓}$$

   Now check. Sum = 4 + (-2) = 2. And -b/a is also 2, so the sum is correct.

5. $$\text{Product}=4\times (-2)=-8={\displaystyle \frac{-8}{1}}\mathrm{✓}$$

   Product = 4 × (-2) = -8. And c/a is also -8, so the product is also correct.

Final answer$$\text{Zeroes: }4,-2$$

Question 1 (ii)

Find the zeroes of $4s^2-4s+1$ and verify the relationship between the zeroes and the coefficients.

Solution:

1. $$4s^2-4s+1=4s^2-2s-2s+1$$

   Split the middle term: -4s = -2s – 2s, because their product is 4s².

2. $$=2s(2s-1)-1(2s-1)=(2s-1{)}^2$$

   Take out the factor. This is the full square of (2s – 1).

3. $$\text{Zeroes: }{\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{2}}$$

   So s = 1/2. Both zeroes are 1/2, so they are repeated zeroes.

4. $$\text{Sum}={\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2}}=1=-{\displaystyle \frac{-4}{4}}\mathrm{✓}$$

   Sum = 1/2 + 1/2 = 1. And -b/a is also 1, so it is correct.

5. $$\text{Product}={\displaystyle \frac{1}{2}}\times {\displaystyle \frac{1}{2}}={\displaystyle \frac{1}{4}}={\displaystyle \frac{1}{4}}\mathrm{✓}$$

   Product = 1/2 × 1/2 = 1/4. And c/a is also 1/4, so it is correct.

Final answer$$\text{Zeroes: }{\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{2}}$$

Question 1 (iii)

Find the zeroes of $6x^2-3-7x$ and verify the relationship between the zeroes and the coefficients.

Solution:

1. $$6x^2-7x-3=6x^2-9x+2x-3$$

   First, write the polynomial in the correct order: 6x² – 7x – 3. Then split the middle term as -9x + 2x.

2. $$=3x(2x-3)+1(2x-3)=(3x+1)(2x-3)$$

   Take out the factors to get (3x + 1)(2x – 3).

3. $$\text{Zeroes: }-{\displaystyle \frac{1}{3}},{\displaystyle \frac{3}{2}}$$

   Setting these to zero gives x = -1/3 and x = 3/2.

4. $$\text{Sum}=-{\displaystyle \frac{1}{3}}+{\displaystyle \frac{3}{2}}={\displaystyle \frac{7}{6}}=-{\displaystyle \frac{-7}{6}}\mathrm{✓}$$

   Sum = -1/3 + 3/2 = 7/6. And -b/a is also 7/6, so it is correct.

5. $$\text{Product}=-{\displaystyle \frac{1}{3}}\times {\displaystyle \frac{3}{2}}=-{\displaystyle \frac{1}{2}}={\displaystyle \frac{-3}{6}}\mathrm{✓}$$

   Product = -1/2. And c/a = -3/6 = -1/2, so it is correct.

Final answer$$\text{Zeroes: }-{\displaystyle \frac{1}{3}},{\displaystyle \frac{3}{2}}$$

Question 1 (iv)

Find the zeroes of $4u^2+8u$ and verify the relationship between the zeroes and the coefficients.

Solution:

1. $$4u^2+8u=4u(u+2)$$

   There is no constant term here, so directly take out 4u to get 4u(u + 2).

2. $$4u=0\text{ or }u+2=0$$

   Set both factors equal to zero.

3. $$\text{Zeroes: }0,-2$$

   So the zeroes are 0 and -2.

4. $$\text{Sum}=0+(-2)=-2=-{\displaystyle \frac{8}{4}}\mathrm{✓}$$

   Sum = -2. And -b/a = -8/4 = -2, so it is correct.

5. $$\text{Product}=0\times (-2)=0={\displaystyle \frac{0}{4}}\mathrm{✓}$$

   Product = 0. And c/a = 0/4 = 0, so it is correct.

Final answer$$\text{Zeroes: }0,-2$$

Question 1 (v)

Find the zeroes of $t^2-15$ and verify the relationship between the zeroes and the coefficients.

Solution:

1. $$t^2-15=t^2-(\sqrt{15}{)}^2$$

   This uses the identity a² – b². Write 15 as the square of √15.

2. $$=(t-\sqrt{15})(t+\sqrt{15})$$

   So the factors are (t – √15)(t + √15).

3. $$\text{Zeroes: }\sqrt{15},-\sqrt{15}$$

   The zeroes are √15 and -√15.

4. $$\text{Sum}=\sqrt{15}-\sqrt{15}=0=-{\displaystyle \frac{0}{1}}\mathrm{✓}$$

   Sum = √15 – √15 = 0. And -b/a is also 0, so it is correct.

5. $$\text{Product}=\sqrt{15}\times (-\sqrt{15})=-15={\displaystyle \frac{-15}{1}}\mathrm{✓}$$

   Product = -15. And c/a is also -15, so it is correct.

Final answer$$\text{Zeroes: }\sqrt{15},-\sqrt{15}$$

Question 1 (vi)

Find the zeroes of $3x^2-x-4$ and verify the relationship between the zeroes and the coefficients.

Solution:

1. $$3x^2-x-4=3x^2+3x-4x-4$$

   Split the middle term: -x = +3x – 4x, because their product is -12x².

2. $$=3x(x+1)-4(x+1)=(3x-4)(x+1)$$

   Take out the factors to get (3x – 4)(x + 1).

3. $$\text{Zeroes: }{\displaystyle \frac{4}{3}},-1$$

   The zeroes are 4/3 and -1.

4. $$\text{Sum}={\displaystyle \frac{4}{3}}+(-1)={\displaystyle \frac{1}{3}}=-{\displaystyle \frac{-1}{3}}\mathrm{✓}$$

   Sum = 4/3 – 1 = 1/3. And -b/a is also 1/3, so it is correct.

5. $$\text{Product}={\displaystyle \frac{4}{3}}\times (-1)=-{\displaystyle \frac{4}{3}}={\displaystyle \frac{-4}{3}}\mathrm{✓}$$

   Product = -4/3. And c/a is also -4/3, so it is correct.

Final answer$$\text{Zeroes: }{\displaystyle \frac{4}{3}},-1$$

Question 2

Question 2 (i)

Find a quadratic polynomial with sum and product of zeroes: ${\displaystyle \frac{1}{4}},-1$

Solution:

1. $$x^2-(\text{sum})x+\text{product}$$

   Use the formula: x² – (sum)x + product.

2. $$x^2-{\displaystyle \frac{1}{4}}x+(-1)$$

   The sum is 1/4 and the product is -1, so we get x² – (1/4)x – 1.

3. $$4x^2-x-4(\times 4)$$

   Multiply by 4 to remove the fractions. The polynomial is 4x² – x – 4.

Final answer$$4x^2-x-4(\times 4)$$

Question 2 (ii)

Find a quadratic polynomial with sum and product of zeroes: $\sqrt{2},{\displaystyle \frac{1}{3}}$

Solution:

1. $$x^2-(\text{sum})x+\text{product}$$

   Use the same formula: x² – (sum)x + product.

2. $$x^2-\sqrt{2}x+{\displaystyle \frac{1}{3}}$$

   The sum is √2 and the product is 1/3, so we get x² – √2x + 1/3.

3. $$3x^2-3\sqrt{2}x+1(\times 3)$$

   Multiply by 3 to get 3x² – 3√2x + 1.

Final answer$$3x^2-3\sqrt{2}x+1(\times 3)$$

Question 2 (iii)

Find a quadratic polynomial with sum and product of zeroes: $0,\sqrt{5}$

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Solution:

1. $$x^2-(\text{sum})x+\text{product}$$

   The formula is x² – (sum)x + product.

2. $$x^2-0\cdot x+\sqrt{5}$$

   The sum is 0, so the x term goes away, and the product is √5.

3. $$x^2+\sqrt{5}$$

   So the polynomial is x² + √5.

Final answer$$x^2+\sqrt{5}$$

Question 2 (iv)

Find a quadratic polynomial with sum and product of zeroes: $1,1$

Solution:

1. $$x^2-(\text{sum})x+\text{product}$$

   Use the formula: x² – (sum)x + product.

2. $$x^2-1\cdot x+1$$

   The sum is 1 and the product is 1, so we get x² – x + 1.

3. $$x^2-x+1$$

   This is our polynomial.

Final answer$$x^2-x+1$$

Question 2 (v)

Find a quadratic polynomial with sum and product of zeroes: $-{\displaystyle \frac{1}{4}},{\displaystyle \frac{1}{4}}$

Solution:

1. $$x^2-(\text{sum})x+\text{product}$$

   The formula is x² – (sum)x + product.

2. $$x^2+{\displaystyle \frac{1}{4}}x+{\displaystyle \frac{1}{4}}$$

   The sum is -1/4, so minus and minus become plus. This gives x² + (1/4)x + 1/4.

3. $$4x^2+x+1(\times 4)$$

   Multiply by 4 to get 4x² + x + 1.

Final answer$$4x^2+x+1(\times 4)$$

Question 2 (vi)

Find a quadratic polynomial with sum and product of zeroes: $4,1$

Solution:

1. $$x^2-(\text{sum})x+\text{product}$$

   The formula is x² – (sum)x + product.

2. $$x^2-4x+1$$

   The sum is 4 and the product is 1, so we get x² – 4x + 1.

3. $$x^2-4x+1$$

   This is our answer. Remember, these questions can have other answers too, which are k times this polynomial.

Final answer$$x^2-4x+1$$

Key Takeaways – Exercise 2.2

• For a quadratic polynomial ax^2 + bx + c, sum of zeroes = -b/a and product of zeroes = c/a.
• You can find zeroes by factorising the quadratic polynomial or by splitting the middle term.
• Given only the sum (S) and product (P) of zeroes, the quadratic polynomial is k(x^2 – Sx + P) for any non-zero constant k.
• Always verify your zeroes by substituting back into the sum and product formulas — this is what ‘verify’ means in Q1.
• Exercise 2.2 has 8 questions (Q1 with 6 parts and Q2 with 6 parts), all based on quadratic polynomials only.

CBSE Board Exam Tips

• In board exams, ‘verify the relationship’ means you must explicitly write: sum of zeroes = alpha + beta = -b/a AND product of zeroes = alpha × beta = c/a — do not skip either formula or you may lose marks.
• For Q2-type questions, always write the general form k(x^2 – Sx + P) first, then substitute k = 1 (or another integer) to get a polynomial with integer/rational coefficients — examiners prefer the simplest form.
• When zeroes involve surds (like sqrt(15) or sqrt(2)), double-check your product: sqrt(15) × (-sqrt(15)) = -15, not +15 — sign errors are the most common mistakes here.
• Rewrite the polynomial in standard form (ax^2 + bx + c) before identifying a, b, c — for example, 6x^2 – 7x – 3 must be rewritten from 6x^2 – 3 – 7x before you apply the formulas.

Practice MCQs – Exercise 2.2

Frequently Asked Questions

How many questions are in NCERT Class 10 Maths Exercise 2.2?

Exercise 2.2 has 2 main questions with a total of 12 parts — Q1 has 6 parts (i to vi) where you find zeroes and verify relationships, and Q2 has 6 parts (i to vi) where you form a quadratic polynomial given its sum and product of zeroes. All 12 parts are based on quadratic polynomials only.

Is Exercise 2.2 of Class 10 Maths important for the CBSE board exam?

Yes, Exercise 2.2 is one of the most important exercises in Chapter 2 and is frequently tested in CBSE board exams. Questions based on finding zeroes, verifying relationships, and forming quadratic polynomials from given zeroes appear almost every year, either as 2-mark or 3-mark questions.

What is the relationship between zeroes and coefficients of a quadratic polynomial?

For a quadratic polynomial ax^2 + bx + c with zeroes alpha and beta, the two key relationships are: sum of zeroes (alpha + beta) = -b/a, and product of zeroes (alpha × beta) = c/a. These formulas are the heart of Exercise 2.2 and must be memorised for the board exam.

How do you form a quadratic polynomial if you are given the sum and product of its zeroes?

If the sum of zeroes is S and the product of zeroes is P, then the quadratic polynomial is given by k(x^2 – Sx + P), where k is any non-zero real constant. For the simplest answer, take k = 1, or choose k to clear fractions and get integer coefficients.

Why does 4u^2 + 8u have 0 as one of its zeroes?

When you factorise 4u^2 + 8u, you get 4u(u + 2), which gives zeroes u = 0 and u = -2. Zero is a perfectly valid zero of a polynomial — it simply means the polynomial equals zero when u = 0, which you can verify by substituting: 4(0)^2 + 8(0) = 0.

Can a quadratic polynomial have two equal (repeated) zeroes? Give an example from Exercise 2.2.

Yes, a quadratic polynomial can have two equal zeroes, also called a repeated root. In Exercise 2.2, Q1(ii) — the polynomial 4s^2 – 4s + 1 = (2s – 1)^2 — has both zeroes equal to 1/2. This happens when the discriminant (b^2 – 4ac) equals zero.

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NCERT Class 10 Maths Solutions – All Exercises

Source: Questions from the NCERT Mathematics Textbook for Class 10, Chapter 2. Solutions written by IrfanEdu.