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

AcademicMathematicsNCERTClass 10

Given: 

Sum and product of the zeroes of the quadratic polynomials are given.

To do: 

we have to find the required quadratic polynomial each with the given numbers .

Solution:

(i)  Let $\alpha$ and $\beta$ are the zeroes of the polynomial.

As given,

Sum of the zeroes$=\alpha+\beta=\frac{1}{4}$

Product of the zeroes$=\alpha.\beta=-1$

The quadratic polynomial is:

$x^{2}-( \alpha +\beta )+\alpha \beta =0$

$\Rightarrow x^2-( \frac{1}{4})x+( -1)=0$

$\Rightarrow 4x^2-x-4=0$

Thus, the required polynomial is $4x^2-x-4=0$. 

(ii) Let $\alpha$ and $\beta$ are the zeroes of the polynomial.

As given, 

Sum of the zeroes$=\alpha +\beta=\sqrt{2}$

Product of the quadratic polynomial$=\alpha.\beta =\frac{1}{3}$

The quadratic polynomial is:

$x^{2}-( \alpha +\beta )+\alpha \beta =0$

$\Rightarrow x^{2}-\sqrt{2}x+\frac{1}{3}=0$

$\Rightarrow 3x^{2}-3\sqrt{2}x+1=0$

Thus, the required quadratic polynomial is $3x^{2}-3\sqrt{2}x+1$.

(iii) Let $\alpha$ and $\beta$ be the zeroes of the quadratic polynomial.

Sum of the zeroes of quadratic polynomial$=\alpha+\beta=0$

Product of the quadratic polynomial$=\alpha.\beta=\sqrt{5}$

The quadratic polynomial is:

$x^{2}-(\alpha +\beta )+\alpha\beta=0$

$\Rightarrow x^{2}-0x+\sqrt{5}=0$

$\Rightarrow x^{2}+\sqrt{5}=0$

The required quadratic polynomial is $x^2+\sqrt{5}$.

(iv) Let $\alpha$ and $\beta$ are the zeroes of the polynomial.

As given, 

Sum of the zeroes$=\alpha +\beta=1$

Product of the zeroes$=\alpha.\beta=1$

$\Rightarrow x^{2}-(\alpha +\beta )+\alpha \beta =0$

$\Rightarrow x^{2}-1.x+1=0$

$\Rightarrow x^{2}-x+1=0$

Thus, the required quadratic polynomial is $x^2-x+1$.

(v) Let $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial.

As given, Sum of the zeroes$=\alpha +\beta=-\frac{1}{4}$

Product of the zeroes$=\alpha \beta =\frac{1}{4}$

$x^{2}-( \alpha+\beta)+\alpha\beta =0$

$x^{2}-( -\frac{1}{4})x+\frac{1}{4}=0$

$4x^{2}+x+1=0$

Thus, he required quadratic polynomial is $4x^2+x+1$

(vi) Let $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial.

As given, sum of the zeroes$=\alpha+\beta=4$

Product of the zeroes$=\alpha\beta=1$

The quadratic polynomial is:

$x^{2}-( \alpha +\beta )+\alpha \beta=0$

$x^{2}-4x+1=0$

The required quadratic polynomial is $x^2-4x+1$.

raja
Updated on 10-Oct-2022 13:19:34

Advertisements