Exercise 2.2

Q 1 (i & ii), Ex 2.2 - Polynomials - Chapter 2 - Maths Class 10th - NCERT

Q 1 (iii & iv), Ex 2.2 - Polynomials - Chapter 2 - Maths Class 10th - NCERT

Q 1 (v & vi), Ex 2.2 - Polynomials - Chapter 2 - Maths Class 10th - NCERT

Question 1
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) $x^2-2x-8$
Sol :
$x^2-2x-8=(x-4)(x+2)$
$x^2-2x-8=0$
$\Rightarrow (x-4)=0$ or $(x+2)=0$
$\Rightarrow x=4$ or $x=-2$
Therefore , the zeroes of $x^2-2x$ are 4 and -2 .
Sum of zeroes $=4-2=2$ $=\dfrac{-(-2)}{1}$ $=\dfrac{-(\text{coefficient of }x)}{~~~\text{cofficient of } x^2}$
Product of zeroes $=4\times (-2)=8$ $=\dfrac{-8}{\phantom{-}1}$ $=\dfrac{\text{constant term}}{\text{cofficient of }x^2}$

(ii) $4s^2-4s+1$
Sol :
$4s^2-4s+1=(2s-1)^2$
$4s^2-4s+1=0$
$\Rightarrow 2s-1=0$
$\Rightarrow s=\dfrac{1}{2}$
Therefore , the zeroes of $4s^2-4s+1$ are $~\dfrac{1}{2}$ and $\dfrac{1}{2}$
Sum of zeroes $=\dfrac{1}{2}+\dfrac{1}{2}=1$ $=\dfrac{-(-4)}{4}$ $=\dfrac{-(\text{coefficient of }s)}{~~~\text{cofficient of } s^2}$
Product of zeroes $=\dfrac{1}{2}\times\dfrac{1}{2}$ $=\dfrac{1}{4}$ $\dfrac{\text{constant term}}{\text{cofficient of }s^2}$

(iii) $6x^2-3-7x$
Sol :
$6x^2-3-7x=(3x+1)(2x-3)$
$6x^2-3-7x=0$
$\Rightarrow 3x+1=0$ or $2x-3=0$
$\Rightarrow x=\dfrac{-1}{\phantom{-}3}$ or $x=\dfrac{3}{2}$
Therefore , the zeroes of $6x^2-3-7x$ are $~\dfrac{-1}{\phantom{-}3}$ and $x=\dfrac{3}{2}$
Sum of zeroes $=\dfrac{-1}{\phantom{-}3}+\dfrac{3}{2}$ $=\dfrac{7}{6}$ $=\dfrac{-(-7)}{6}$ $=\dfrac{-(\text{coefficient of }x)}{~~~\text{cofficient of } x^2}$
Product of zeroes $=\dfrac{-1}{\phantom{-}3}\times\dfrac{3}{2}$ $=\dfrac{-1}{\phantom{-}2}$ $=\dfrac{-3}{\phantom{-}6}$ $=\dfrac{\text{constant term}}{\text{cofficient of }x^2}$

(iv) $4u^2+8u$
Sol :
$4u^2+8u=(4u)(u+2)$
$4u^2+8u=0$
$\Rightarrow (4u)=0$ or $(u+2)=0$
$\Rightarrow u=0$ or $u=-2$
Therefore , the zeroes of $4u^2+8u$ are 0 and - 2 .
Sum of zeroes $=0+(-2)=-2$ $=\dfrac{-(-8)}{4}$ $=\dfrac{-(\text{coefficient of }u)}{~~~\text{cofficient of } u^2}$
Product of zeroes $=0\times (-2)=0$ $=\dfrac{0}{4}$ $=\dfrac{\text{constant term}}{\text{cofficient of }u^2}$

(v) $t^2-15$
Sol :
$t^2-15=(t+\sqrt{15})(t-\sqrt{15})$ $[\because a^2-b^2=(a+b)(a-b)]$
$t^2-15=0$
$\Rightarrow (t+\sqrt{15})=0$ or $(t-\sqrt{15})=0$
$\Rightarrow t=-\sqrt{15}$ or $t=\sqrt{15}$
Therefore , the zeroes of $t^2-15$ are $t=-\sqrt{15}$ and $t=+\sqrt{15}$
Sum of zeroes $=\sqrt{15}+(-\sqrt{15})=0$ $=\dfrac{-(0)}{1}$ $=\dfrac{-(\text{coefficient of }t)}{~~~\text{cofficient of } t^2}$
Product of zeroes $=\sqrt{15}\times(-\sqrt{15})$ $=-15$=\dfrac{-15}{1}$ $=\dfrac{\text{constant term}}{\text{cofficient of }t^2}$

(vi) $3x^2-x-4$
Sol :
$3x^2-x-4=(3x-4)(x+1)$
$3x^2-x-4=0$
$\Rightarrow (3x-4)=0$ or $(x+1)=0$
$\Rightarrow x=\dfrac{4}{3}$ or $x=-1$
Therefore , the zeroes of $3x^2-x-4$ are $~\dfrac{4}{3}$ and $x=-1$
Sum of zeroes $=\dfrac{4}{3}+(-1)$ $=\dfrac{1}{3}$ $=\dfrac{-(-1)}{3}$ $=\dfrac{-(\text{coefficient of }x)}{~~~\text{cofficient of } x^2}$
Product of zeroes $=\dfrac{4}{3}\times(-1)$ $=\dfrac{-4}{\phantom{-}3}$ $=\dfrac{\text{constant term}}{\text{cofficient of }x^2}$

Concept insight: The zero of a polynomial is that value of the variable which when substituted in the polynomial makes its value 0.
When a quadratic polynomial is equated to 0, then the values of the variable obtained are the zeroes of that polynomial. The relationship between the zeroes of a quadratic polynomial with its coefficients is very important. Also, while verifying the above relationships, be careful about the signs of the coefficients.

Q 2 (i,ii & iii), Ex 2.2 - Polynomials - Chapter 2 - Maths Class 10th - NCERT

Q 2 (iv,v & vi), Ex 2.2 - Polynomials - Chapter 2 - Maths Class 10th - NCERT

(working)
Question 2
Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) $\dfrac{1}{4},1$
(ii) $\sqrt{2},\dfrac{1}{3}$
(iii) $0,\sqrt{5}$
(iv) 1 , 1
(v) $-\dfrac{1}{4},\dfrac{1}{4}$
(vi) 4 , 1