Chapter 2: Polynomials

Q&A -Ask Doubts and Get Answers

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

Answer:

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 .

VIDEO EXPLANATION


Related Questions for Study

What our students and parents say about us!

Choose EduSakshamยฎ
Embrace Better Learning

edusaksham-mobile-app