Question

If we are given $\sqrt{5}+\sqrt{2}$ is the root of $3{{x}^{5}}-4{{x}^{4}}-42{{x}^{3}}+56{{x}^{2}}+27x-36=0$, then the rational root is:
a). $\dfrac{4}{3}$
b). $\dfrac{3}{4}$
c). $\dfrac{1}{4}$
d). $-\dfrac{4}{3}$

Answer 1

Hint: To solve the question we will use the concept of conjugates roots of any equation to find all the possible irrational roots of the given equation and then to find the rational root we will use the concept that if $a{{x}^{5}}+b{{x}^{4}}+c{{x}^{3}}+d{{x}^{2}}+ex+f=0$ is an equation whose roots are v, w, x, y, z. Then, we know that the sum of roots is equal to $-\dfrac{b}{a}$ $\Rightarrow v+w+x+y+z=-\dfrac{b}{a}$

Complete step-by-step solution
Let us assume a quadratic equation $a{{x}^{2}}+bx+c=0$ and let us also say that $a+\sqrt{b}$ is one of its roots, then we will say that its other root must be its conjugate i.e. $a-\sqrt{b}$.
Since, we know that $\sqrt{5}+\sqrt{2}$ is the root of $3{{x}^{5}}-4{{x}^{4}}-42{{x}^{3}}+56{{x}^{2}}+27x-36=0$, and we can see that $\sqrt{5}+\sqrt{2}$is irrational so we also know that irrational roots occur in a conjugates.
So, all possible conjugates pairs of $\sqrt{5}+\sqrt{2}$ are $\sqrt{5}-\sqrt{2}$, $-\sqrt{5}+\sqrt{2}$, $-\sqrt{5}-\sqrt{2}$ because in $\sqrt{5}+\sqrt{2}$ both the terms (i.e.$\sqrt{5}$ and $\sqrt{2}$) are irrational.
So, the roots of the equation $3{{x}^{5}}-4{{x}^{4}}-42{{x}^{3}}+56{{x}^{2}}+27x-36=0$ are all the possible conjugates of $\sqrt{5}+\sqrt{2}$,
Hence, $\sqrt{5}-\sqrt{2}$, $-\sqrt{5}+\sqrt{2}$, $-\sqrt{5}-\sqrt{2}$ is also the roots of the equation $3{{x}^{5}}-4{{x}^{4}}-42{{x}^{3}}+56{{x}^{2}}+27x-36=0$ along with $\sqrt{5}+\sqrt{2}$.
Since, equation $3{{x}^{5}}-4{{x}^{4}}-42{{x}^{3}}+56{{x}^{2}}+27x-36=0$ is an equation of degree 5 so there must be 5 possible roots of the above equation, and four of them are $\sqrt{5}+\sqrt{2}$, $\sqrt{5}-\sqrt{2}$, $-\sqrt{5}+\sqrt{2}$, $-\sqrt{5}-\sqrt{2}$ which are all irrational.
Now, only one root of the equation $3{{x}^{5}}-4{{x}^{4}}-42{{x}^{3}}+56{{x}^{2}}+27x-36=0$ is remaining to be found so it must be rational because irrational roots occur in conjugates.
Let us say that $a{{x}^{5}}+b{{x}^{4}}+c{{x}^{3}}+d{{x}^{2}}+ex+f=0$ is equation whose roots are v, w, x, y, z .
Then, we know that the sum of roots is equal to $-\dfrac{b}{a}$.
$\Rightarrow v+w+x+y+z=-\dfrac{b}{a}$
So, for equation $3{{x}^{5}}-4{{x}^{4}}-42{{x}^{3}}+56{{x}^{2}}+27x-36=0$, the sum of roots = $-\dfrac{(-4)}{3}$
Let us say that $\alpha $ is the rational root of the equation $3{{x}^{5}}-4{{x}^{4}}-42{{x}^{3}}+56{{x}^{2}}+27x-36=0$,
So, sum of roots = $\left( \sqrt{5}+\sqrt{2} \right)$ + $\left( \sqrt{5}-\sqrt{2} \right)$ + $\left( -\sqrt{5}+\sqrt{2} \right)$ + $\left( -\sqrt{5}-\sqrt{2} \right)$ + $\alpha $ = $-\dfrac{(-4)}{3}$
So, $\alpha $ = $\dfrac{4}{3}$
Hence, option (a) is our correct answer.

Note: Students are required to keep in mind that if we have been given a root of an equation and the root is irrational then all the possible conjugates of that root will be the roots of the given equation. Also, students are required to note that if we have an equation whose degree is n i.e.${{a}_{0}}{{x}^{n}}+{{a}_{1}}{{x}^{n-1}}+{{a}_{2}}{{x}^{n-2}}+{{a}_{3}}{{x}^{n-3}}+..........+{{a}_{n-3}}{{x}^{3}}{{a}_{n-2}}{{x}^{2}}{{a}_{n-1}}x+{{a}_{n}}=0$ then the sum of all the roots is equal to $\dfrac{-{{a}_{1}}}{{{a}_{0}}}$ and the products of the roots is equal to ${{\left( -1 \right)}^{n}}\dfrac{{{a}_{n}}}{{{a}_{0}}}$.