Question

Find the quadratic polynomial, for the zeroes a, ß given in each case.
(i) 1/4,-1

(ii) 13,-13
(iii)
A
-1
(iv)
1 3
2' 2​

Answer 1

x^2-3

(iii)4x^2+3x-1

(iv)4x^2-8x+3

Step-by-step explanation:

(i)We are given that the zeroes of quadratic polynomial

\alpha=2,\bet=-1

We have to find the quadratic polynomial

We know that general quadratic polynomial

x^2-(\alpha+\beta)x+ \alpha\times \beta

Substituting the values of zeroes in general polynomial

Then we get x^2-(2-1)x+2\times(-1)

x^2-x-2

Hence, the quadratic polynomial is given by

x^2-x-2

(ii)We are given that zeroes of quadratic polynomial

\alpha=\sqrt3,\beta=-\sqrt3

Substituting the values of zeroes in the general quadratic polynomial

Then we have x^2-(\sqrt3-\sqrt3)x-\sqrt3\times \sqrt3

x^2-3

Therefore, the quadratic polynomial

x^2-3

(iii) We are given that two zeroes of quadratic polynomial

\alpha=\frac{1}{4},\beta=-1

Substituting the values of zeroes in the general quadratic polynomial

Then we have x^2-(\frac{1}{4}-1)x-\frac{1}{4}

x^2-\frac{1-4}{4}x-\frac{1}{4}

Substituting the equation equal to zero

x^2+\frac{3}{4}x-\frac{1}{4}=0

\frac{4x^2+3x-1}{4}=0

Hence, the quadratic polynomial

4x^2+3x-1

(iv) We are given that two zeroes of quadratic polynomial

\alpha=\frac{1}{2},\bet=\frac{3}{2}

Substituting the values of zeroes in the general quadratic polynomial

Then we have

x^2-(\frac{1}{2}+\frac{3}{2})x+\frac{1}{2}\times\frac{3}{2}

x^2-\frac{1+3}{2}x+\frac{3}{4}

Substituting equation equal to zero

x^2-2x+\frac{3}{4}=0

\frac{4x^2-8x+3}{4}=0

4x^2-8x+3=0

Hence, the quadratic polynomial

4x^2-8x+3