Question

find quadratic equation for following roots alpha=√3,beta=2√3​

Answer 1

$\mathfrak{\underline{\underline{\green{Answer:-}}}}$

$\mathsf{ {x}^{2}-3\sqrt{3}x + 6}$

$\mathfrak{\underline{\underline{\green{Explanation:-}}}}$

Given:

α and β are the zeros of a quadratic polynomial

$\alpha = \sqrt{3}$

$\beta = 2\sqrt{3}$

To Find:

Quadratic polynomial with the given zeros

Solution:

Let us find the sum and product of the zeros of the quadratic polynomial:

Sum

$\mathsf{ = \alpha + \beta}$

$\mathsf{ = \sqrt{3} + 2\sqrt{3} }$

$\mathsf{ = 3\sqrt{3} }$

Product

$\mathsf{ = \alpha \times \beta}$

$\mathsf{ = \sqrt{3} \times 2\sqrt{3}}$

$\mathsf{ = 6}$

From Quadratic formula

$\boxed{\pink{ {x}^{2}-(\alpha + \beta)x + \alpha \beta}}$

By sub. the values

$\mathsf{ ={x}^{2}-( 3\sqrt{3})x + 6}$

$\mathsf{ ={x}^{2}-3\sqrt{3}x + 6}$

Hence

$\mathsf{ {x}^{2}-3\sqrt{3}x + 6}$ is the required quadratic polynomial

Answer 2

ANSWER

given

$\alpha = \sqrt{3} \\ \\ \beta = 2 \sqrt{3}$

let us consider the roots

as 'a' and 'b'

REQUIRED TO FIND:

the quadratic equation of the roots

METHOD:

when the roots are given

the formula for the quadratic equation Is

$\sf \boxed{x^2-(a+b)x+ab}$

for this we need to find product and sum of roots

PRODUCT OF ROOTS :

$\sf ab=\sqrt {3}.2\sqrt {3}$

$\sf \underline {ab=6}$

SUM OF ROOTS :

$\sf a+b=\sqrt {3}+2\sqrt{3}$

$\sf \underline {a+b}=3\sqrt{3}$

then the required quadratic equation is

$= > \sf \boxed {{x}^{2} - 3 \sqrt{3} x + 6 = 0}$

IDENTITIES USED:

$= > quadratic \: equation \: = {x}^{2} - (a + b)x + ab$

$= > product \: of \: roots \: = \dfrac{c}{a}=\dfrac {constant \:term}{co\:fficient \: of x^2}$

$= > sum \: of \: roots \: = \dfrac{ - b}{a} =\dfrac {-co \: efficient \: of \:x}{co \: efficient \: of \: x^2}$