NCERT solution class 10 chapter 2 polynomials

Exercise 2.2 (working)

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[latex]=\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.

 

(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

 

 

 

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