Question

2 points
10) If A and B are the zero of the
quadratic polynomial x2- 25 then
the quadratic polynomial whose
zeros are (1+A) and (1 + B) is
O X2 - 24x + 2
O X2 - 2x + 24
O X2 - 2x - 24
O
X2 + 2x + 24​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:A \: and \: B \: are \: zeroes \: of \: {x}^{2} - 25$

We know that

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\bf\implies \:A + B = - \dfrac{0}{1} = 0 - - - (1)$

And

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\bf\implies \:AB = \dfrac{ - 25}{1} = - 25 - - - (2)$

Now,

We have to find a polynomial whose zeroes are 1 + A and 1 + B.

Consider,

Sum of the zeroes,

$\rm :\longmapsto\:S = 1 + A + 1 + B$

$\: \rm \: = \: \:2 + A + B$

$\: \rm \: = \: \:2 + 0$

$\: \rm \: = \: \:2$

$\bf\implies \:S = 2 - - - (3)$

Consider,

Product of zeroes,

$\rm :\longmapsto\:P = (1 + A)(1 + B)$

$\: \rm \: = \: \:1 + A + B + AB$

$\: \rm \: = \: \:1 + 0 - 25$

$\: \rm \: = \: \: - 24$

$\bf\implies \:P = - 24 - - - (4)$

Thus,

The required Quadratic polynomial is

$\rm :\longmapsto\:f(x) = k\bigg( {x}^{2} - Sx + P \bigg) \: where \: k \: \ne \: 0$

$\rm :\longmapsto\:f(x) = k\bigg( {x}^{2} - 2x - 24 \bigg) \: where \: k \: \ne \: 0$

Hence,

$\red{\bf :\longmapsto\:f(x) = {x}^{2} - 2x - 24}$