Question

frame a quadratic equation in x when roots are (2+√5) and (2-√5)

Answer 1

Given:-

Roots of the quadratic equation = (2+√5) and (2-√5)

To find:-

The quadratic equation.

Assumption:-

Let $\alpha$ and $\beta$ be two zeroes of the required quadratic equation.

$\sf{\alpha}$ = (2+√5)

$\sf{\beta}$ = (2-√5)

Solution:-

We know,

A quadratic equation is always in the form,

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

Putting the values,

$\sf{x^2 - [(2+\sqrt{5}) + (2-\sqrt{5})]x + [(2+\sqrt{5})(2-\sqrt{5})]}$

= $\sf{x^2 - [2+\sqrt{5} + 2-\sqrt{5}]c + [(2)^2 - (\sqrt{5})^2]}$

$\sf{\because (a+b)(a-b) = a^2 - b^2}$

= $\sf{x^2 - [4 + \cancel{\sqrt{5}} - \cancel{\sqrt{5}}]x + [4-5]}$

= $\sf{x^2 - (4)x + (-1)}$

= $\sf{x^2 - 4x - 1}$

Hence the required quadratic equation is x² - 4x - 1.

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Verification:-

Let us verify that the quadratic equation we got is correct or not.

We know,

Sum of zeroes = $\sf{\dfrac{-Coefficient\:of\:x}{Coefficient\:of\:x^2}}$

= $\sf{(2+\sqrt{5}) + (2-\sqrt{5}) = \dfrac{-(-4)}{1}}$

= $\sf{2+\sqrt{5} + 2 - \sqrt{5} = 4}$

= $\sf{4 = 4}$

Product of zeroes = $\sf{Constant\:Term}{Coefficient\:of\:x^2}}$

= $\sf{(2+\sqrt{5})(2-\sqrt{5}) = \dfrac{-1}{1}}$

$\sf{\because (a+b)(a-b) = a^2 - b^2}$

= $\sf{(2)^2 - (\sqrt{5})^2 = -1}$

= $\sf{4-5 = -1}$

= $\sf{-1 = -1}$

Hence Verified!!!

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