Question

14 .The quadratic equation whose one root is 2-root3 is​

Answer 1

Answer:

$x^{2}-4x +1 = 0$

Step-by-step explanation:

Whenever one root is given in form of (Integer)-(irrational number) then the other root is sure to be (Integer)+(irrational number)

Thus, (2-√3) is one root , the other root is (2+√3)

Now, we have both the roots, α=(2+√3) and β=(2-√3)

Form of equation when we have both the roots is ==>

$x^{2}-(\alpha+\beta )x +\alpha.\beta = 0$

Now,

α+β = (2+√3) + (2-√3) = 4

α.β = (2+√3)(2-√3) = [(2)^2 - (√3)^2]......[(a+b)(a-b) = (a)^2 - (b)^2]

α.β = 4-3 = 1

Thus

$x^{2}-(\alpha+\beta )x +\alpha.\beta = 0$

==> $x^{2}-4x +1 = 0$    ........(To be found)

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