### Chapter 2: Polynomials

Q
##### Polynomials CBSE NCERT Solutions

Question:

Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

(i) 1/4  , 1                                        (ii) $\sqrt 2$  , 1/3                                  (iii) 0, $\sqrt 5$

(iv) 1, 1                                          (v) -1/4,  1/4                                 (vi) 4,1

Solutions:

(i) 1/4  , 1

Given, the sum of zeroes  $\alpha + \beta$  is 1/4   and the product of zeroes   $\alpha \times \beta$ is 1.

We know that the quadratic equation formula is $x^{2}$  – (sum of the root)  + (product of root) = 0.

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

Substitute the given values in the formula, we get

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

$\Rightarrow 4x^{2}-x + 4 = 0$

Hence, the quadratic polynomial is  $4x^{2}-x + 4$.

(ii) $\sqrt 2$  , 1/3

Given, the sum of zeroes  $\alpha + \beta$  is $\sqrt 2$   and the product of zeroes   $\alpha \times \beta$ is 1/3.

We know that the quadratic equation formula is $x^{2}$  – (sum of the root)  + (product of root) = 0.

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

Substitute the given values in the formula, we get

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

$\Rightarrow 3x^{2}-3\sqrt2 x + 1 = 0$

Hence, the quadratic polynomial is  $3x^{2}-3\sqrt2 x + 1$.

(iii) 0, $\sqrt 5$

Given, the sum of zeroes  $\alpha + \beta$  is 0   and the product of zeroes   $\alpha \times \beta$ is $\sqrt 5$.

We know that the quadratic equation formula is $x^{2}$  – (sum of the root)  + (product of root) = 0.

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

Substitute the given values in the formula, we get

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

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

Hence, the quadratic polynomial is  $x^{2} + \sqrt{5}$.

(iv) 1, 1

Given, the sum of zeroes  $\alpha + \beta$  is 1   and the product of zeroes   $\alpha \times \beta$ is 1.

We know that the quadratic equation formula is $x^{2}$  – (sum of the root)  + (product of root) = 0.

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

Substitute the given values in the formula, we get

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

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

Hence, the quadratic polynomial is  $x^{2} - x +1$ .

(v) -1/4 1/4

Given, the sum of zeroes  $\alpha + \beta$  is -1/4   and the product of zeroes   $\alpha \times \beta$ is 1/4.

We know that the quadratic equation formula is $x^{2}$  – (sum of the root)  + (product of root) = 0.

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

Substitute the given values in the formula, we get

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

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

Hence, the quadratic polynomial is  $4x^{2} + x +1$ .

(vi) 4,1

Given, the sum of zeroes  $\alpha + \beta$  is 4   and the product of zeroes   $\alpha \times \beta$ is 1.

We know that the quadratic equation formula is $x^{2}$  – (sum of the root)  + (product of root) = 0.

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

Substitute the given values in the formula, we get

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

Hence, the quadratic polynomial is  $x^{2} - 4x +1$ .

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