Question

Q11. Find the quadratic polynomial whose zeroes are 5 + √3 and 5 - √3.​

Answer 1

Given :-

• Zeroes of a quadratic polynomial are 5 + √3 and 5 - √3.

To Find :-

• The quadratic polynomial.

Solution :-

Let the zeroes of quadratic polynomial is represented as

$\rm :\longmapsto\: \alpha = 5 + \sqrt{3}$

$\rm :\longmapsto\: \beta = 5 - \sqrt{3}$

So,

Sum of the zeroes is

$\rm :\longmapsto\: \alpha + \beta$

$\rm \: = \: 5 + \sqrt{3} + 5 - \sqrt{3}$

$\rm \: = \: 10$

$\bf\implies \:\red{\boxed{\sf \: \alpha + \beta = 10}}$

Now,

Product of zeroes is

$\rm :\longmapsto\: \alpha \beta$

$\rm \: = \: (5 + \sqrt{3})(5 - \sqrt{3})$

$\rm \: = \: {5}^{2} - {( \sqrt{3}) }^{2}$

$\rm \: = \: 25 - 3$

$\rm \: = \: 22$

$\bf\implies \:\red{\boxed{\sf \: \alpha \beta = 22}}$

So,

Required quadratic polynomial f(x) is given by

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

$\rm :\longmapsto\:f(x) = k( {x}^{2} -10x + 22) \: where \: k \ne \: 0$

Additional Information :-

$\rm \:If \: \alpha \: and \: \beta \: are \: zeroes \: of \: f(x) = {ax}^{2} + bx + c \: then$

$\red{\boxed{\sf \: \alpha + \beta = - \: \dfrac{b}{a}}}$

$\red{\boxed{\sf \: \alpha \beta = \: \dfrac{c}{a}}}$

$\red{\boxed{\sf \: { \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta) }^{2} - 2 \alpha \beta }}$

$\red{\boxed{\sf \: { \alpha }^{3} + { \beta }^{3} = {( \alpha + \beta) }^{3} - 3 \alpha \beta ( \alpha + \beta )}}$

$\red{\boxed{\sf \: | \alpha - \beta | = \sqrt{ {( \alpha + \beta }^{2}) - 4 \alpha \beta}}}$