Question

from the quadratic equation from the equation root given below. 2- root5 , 2+ root5​

Answer 1

Answer:

The answer is $x^2-4x-1=0$

Step-by-step explanation:

Concept: Quadratic equation whose roots are given is given by

                      $x^2-(\text{ sum of roots }) x+\text{ product of roots } =0$

Given roots of the quadratic are $2-\sqrt 5$ and $2+\sqrt 5$

So sum of roots $=2-\sqrt 5+2+\sqrt 5=4$

and product of roots $=(2-\sqrt 5)(2-\sqrt 5)=2^2-(\sqrt 5)^2=4-5=-1$

Therefore the required quadratic equation is,

                   $x^2-(4)x+(-1)=0\Rightarrow x^2-4x-1=0$

Answer 2

QE = $x^{2} -(\alpha+ \beta )x + \alpha \beta = 0$, where α, β are the roots.

α = $2-\sqrt{5}$

β = $2+\sqrt5$

α + β =  $2-\sqrt5+2+\sqrt5$

        = 4

αβ = $(2-\sqrt5)(2+\sqrt5)$                    

    = $2^2 -(\sqrt5)^2$                                     [∵$(a+b)(a-b) = a^2-b^2$]

    = 4 - 5 = -1

∴QE is $x^2 - 4x -1 = 0$