Question

Find the quadratic polynomial, for the zeros α, β given in each case.
(i) 2, –1 (ii) √3, –√3 (iii) 1/4, – 1 (iv) 1/2, 3/2

Answer 1

You can find your answer in the image .

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Answer 2

Answer:

(i)$x^2-x-2$

(ii)$x^2-3$

(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

$4x^2-8x+3$