Question

If $α$ and $β$ are the zeroes of polynomial $x^{2} + 8 x + 6$ . Form a polynomial whose zeroes are $α - β and α + β$ .

Accepted Answer

Given: $α$ and $β$ are the zeroes of polynomial $x^{2} + 8 x + 6$ .

Let $p (x) = a_{2} x^{2} + b_{2} x + c_{2}$ be the polynomial whose zeroes are $α - β and α + β$ .

Here, the coefficients of the polynomial $x^{2} + 8 x + 6$ are $a_{1} = 1, b_{1} = 8, c_{1} = 6$ .

Using the relation between zeroes and coefficients, we can write,

$α + β = \frac{- b_{1}}{a_{1}} = \frac{- 8}{1} = - 8$ ...(i)
$α β = \frac{c_{1}}{a_{1}} = \frac{6}{1} = 6$ ...(ii)

Using (i) and (ii), we can write,

$α + \frac{6}{α} = - 8 [∵ α β = 6] \Rightarrow α^{2} + 6 = - 8 α \Rightarrow α^{2} + 8 α + 6 = 0 \Rightarrow α = \frac{- 8 \pm \sqrt{8^{2} - 4 (6)}}{2} [∵ x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}] = \frac{- 8 \pm 2 \sqrt{10}}{2}$
So, we have $α = \frac{- 8 \pm 2 \sqrt{10}}{2}$ .

We find the sum of zeroes of the polynomial $p (x)$ ,

$α - β + α + β = 2 α = 2 \times \frac{- 8 \pm 2 \sqrt{10}}{2}$
$= - 8 \pm 2 \sqrt{10}$
$\Rightarrow \frac{- 8 \pm 2 \sqrt{10}}{1} = \frac{- b_{2}}{a_{2}}$

Product of zeroes of $p (x) = (α - β) (α + β)$

$= (α + β) \sqrt{{(α + β)}^{2} - 4 α β} = (- 8) \sqrt{{(- 8)}^{2} - 4 (6)} = (- 8) \sqrt{64 - 24} = (- 8) \sqrt{40} = (- 8) (2 \sqrt{10}) = - 16 \sqrt{10}$
$\Rightarrow \frac{- 16 \sqrt{10}}{1} = \frac{c_{2}}{a_{2}}$

Here, the coefficients of $p (x)$ are $a_{2} = 1, b_{2} = 8 \mp 2 \sqrt{10}, c_{2} = - 16 \sqrt{10}$

The polynomial $p (x) = a_{2} x^{2} + b_{2} x + c_{2} = x^{2} + (8 \mp 2 \sqrt{10}) x - 16 \sqrt{10}$

Hence, the other polynomials are $p (x) = k (x^{2} + (8 \mp 2 \sqrt{10}) x - 16 \sqrt{10})$ .

H K Dass, Rama Verma and, Bhagwat Swarup Sharma Solutions for Chapter: Polynomials, Exercise 3: EXERCISE

H K Dass Mathematics Solutions for Exercise - H K Dass, Rama Verma and, Bhagwat Swarup Sharma Solutions for Chapter: Polynomials, Exercise 3: EXERCISE

Questions from H K Dass, Rama Verma and, Bhagwat Swarup Sharma Solutions for Chapter: Polynomials, Exercise 3: EXERCISE with Hints & Solutions

Learn and Prepare for any exam you want !