Class 10 Maths - Chapter 2: Polynomials (Exercise 2.2) - Digital Pipal Academy

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Exercise 2.2

1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

   (i) $x^2 - 2x - 8$
   (ii) $4s^2 - 4s + 1$
   (iii) $6x^2 - 3 - 7x$
   (iv) $4u^2 + 8u$
   (v) $t^2 - 15$
   (vi) $3x^2 - x - 4$

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(i) $x^2 - 2x - 8$

Factorizing:

$$x^2 - 4x + 2x - 8$$ $$x(x - 4) + 2(x - 4)$$ $$(x - 4)(x + 2)$$

Zeroes of the polynomial are $4$ and $-2$.

$$\text{Sum of zeroes} = 4 + (-2) = 2$$ $$= \frac{-(-2)}{1} = \frac{- \text{Coefficient of } x}{\text{Coefficient of } x^2}$$ $$\text{Product of zeroes} = 4 \times (-2) = -8$$ $$= \frac{-8}{1} = \frac{\text{Constant term}}{\text{Coefficient of } x^2}$$

✅ Verified.

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(ii) $4s^2 - 4s + 1$

Factorizing:

$$4s^2 - 2s - 2s + 1$$ $$2s(2s - 1) - 1(2s - 1)$$ $$(2s - 1)(2s - 1)$$

Zeroes of the polynomial are $\frac{1}{2}$ and $\frac{1}{2}$.

$$\text{Sum of zeroes} = \frac{1}{2} + \frac{1}{2} = 1$$ $$= \frac{-(-4)}{4} = \frac{- \text{Coefficient of } s}{\text{Coefficient of } s^2}$$ $$\text{Product of zeroes} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$$ $$= \frac{1}{4} = \frac{\text{Constant term}}{\text{Coefficient of } s^2}$$

✅ Verified.

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(iii) $6x^2 - 7x - 3$

Factorizing:

$$6x^2 - 9x + 2x - 3$$ $$3x(2x - 3) + 1(2x - 3)$$ $$(3x + 1)(2x - 3)$$

Zeroes of the polynomial are $\frac{3}{2}$ and $-\frac{1}{3}$.

$$\text{Sum of zeroes} = \frac{3}{2} + (-\frac{1}{3}) = \frac{7}{6}$$ $$= \frac{-(-7)}{6} = \frac{- \text{Coefficient of } x}{\text{Coefficient of } x^2}$$ $$\text{Product of zeroes} = \frac{3}{2} \times -\frac{1}{3} = -\frac{1}{2}$$ $$= \frac{-3}{6} = \frac{\text{Constant term}}{\text{Coefficient of } x^2}$$

✅ Verified.

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(iv) $4u^2 + 8u$

Factorizing:

$$4u(u + 2)$$

Zeroes of the polynomial are $0$ and $-2$.

$$\text{Sum of zeroes} = 0 + (-2) = -2$$ $$= \frac{-8}{4} = \frac{- \text{Coefficient of } u}{\text{Coefficient of } u^2}$$ $$\text{Product of zeroes} = 0 \times (-2) = 0$$ $$= \frac{0}{4} = \frac{\text{Constant term}}{\text{Coefficient of } u^2}$$

✅ Verified.

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(v) $t^2 - 15$

Factorizing:

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

Zeroes of the polynomial are $\sqrt{15}$ and $-\sqrt{15}$.

$$\text{Sum of zeroes} = \sqrt{15} + (-\sqrt{15}) = 0$$ $$= \frac{-0}{1} = \frac{- \text{Coefficient of } t}{\text{Coefficient of } t^2}$$ $$\text{Product of zeroes} = \sqrt{15} \times (-\sqrt{15}) = -15$$ $$= \frac{-15}{1} = \frac{\text{Constant term}}{\text{Coefficient of } t^2}$$

✅ Verified.

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(vi) $3x^2 - x - 4$

Factorizing:

$$3x^2 - 4x + 3x - 4$$ $$x(3x - 4) + 1(3x - 4)$$ $$(3x - 4)(x + 1)$$

Zeroes of the polynomial are $\frac{4}{3}$ and $-1$.

$$\text{Sum of zeroes} = \frac{4}{3} + (-1) = \frac{1}{3}$$ $$= \frac{-(-1)}{3} = \frac{- \text{Coefficient of } x}{\text{Coefficient of } x^2}$$ $$\text{Product of zeroes} = \frac{4}{3} \times (-1) = -\frac{4}{3}$$ $$= \frac{-4}{3} = \frac{\text{Constant term}}{\text{Coefficient of } x^2}$$

✅ Verified.

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