Question

Find the quadratic equation whose one root is $\dfrac{1}{2+\sqrt{5}}$.

Answer 1

Hint: We need to find the other root to find the quadratic equation. Recall that, if one root of a polynomial is irrational then the other root will be the conjugate of the first root if the polynomial has real coefficients.


Complete step by step solution:We have one root of the quadratic equation. To find the conjugate of the root we need to first rationalize the denominator of the given root.
$\dfrac{1}{2+\sqrt{5}}=\dfrac{2-\sqrt{5}}{(2+\sqrt{5})(2-\sqrt{5})}$
$\dfrac{1}{2+\sqrt{5}}=\dfrac{2-\sqrt{5}}{2^{2}-(\sqrt{5}) 2}$
$\dfrac{1}{2+\sqrt{5}}=-(2-\sqrt{5})$
$\dfrac{1}{2+\sqrt{5}}=-2+\sqrt{5}$
Therefore, the other root will be the conjugate of $-2+\sqrt{5}$, which is equal to $-2-\sqrt{5}$.
Now, we have got the two roots of the required polynomial – $-2+\sqrt{5}$and $-2-\sqrt{5}$.
To find the required polynomial we have,
$x^{2}-(\text { sum of the roots }) x+\text { product of the roots }=0$
$x^{2}-(-2+\sqrt{5}+(-2-\sqrt{5})) x+[(-2+\sqrt{5}) \times(-2-\sqrt{5})]$
$x^{2}-(-4) x+\left((-2)^{2}-(\sqrt{5})^{2}\right)$
$x^{2}+4 x-1$
Therefore, the required quadratic equation is $x^{2}+4 x-1=0$



Option ‘D’ is correct

Note: Whenever we are dealing with irrational numbers it is good to rationalize the denominator so that all the further calculations can be done easily.
- We solved this question by assuming that the quadratic equation is having real coefficients and that is the reason behind taking the conjugate root of one root to be the other root.