Question

If one real root of the quadratic equation $81x^{2}+kx+256=0$ is cube of the other root, then a value of k is
A. -81
B. 100
C. -300
D. 144

Answer 1

Hint: To solve the above equation, we have to assume a variable ‘$r$’ as one of the roots. It is given that one of the roots is the cube root of the other, so the other root becomes ${{r}^{3}}$. Use the equation for finding the sum of roots and the product of roots of the quadratic equation. That is, if the quadratic equation is \[a{{x}^{2}}+bx+c=0\], then its sum of root is $\dfrac{-b}{a}$ and product of root is $\dfrac{c}{a}$.

Complete step by step answer:
Let us assume that one of the roots of the given quadratic equation is $r$.
Now, it is given that one root of the given equation is the cube root of the other. So the other root will be ${{r}^{3}}$.
Now, we have to use the equation to find the sum of roots and product of roots of a quadratic equation.
If the equation is $a{{x}^{2}}+bx+c=0$ then,
Sum of root $=\dfrac{-b}{a}$
Product of root = $\dfrac{c}{a}$
So, for the equation $81{{x}^{2}}+kx+256=0$, we can write,
Sum of root = $\dfrac{-k}{81}\Rightarrow r+{{r}^{3}}=\dfrac{-k}{81}$
And the product of root = $\dfrac{256}{81}$
That is, $r\times {{r}^{3}}=\dfrac{256}{81}\Rightarrow {{r}^{4}}=\dfrac{256}{81}$
We know that $256=4\times 4\times 4\times 4$and $81=3\times 3\times 3\times 3$.
So, we can write, ${{r}^{4}}={{\left( \dfrac{4}{3} \right)}^{4}}$
Therefore, $r=\pm \dfrac{4}{3}$
Now, substitute the value of $r$in the equation $r+{{r}^{3}}=\dfrac{-k}{81}$
Case 1: If $r=\dfrac{4}{3}$ the equation will be,
$\dfrac{4}{3}+{{\left( \dfrac{4}{3} \right)}^{3}}=\dfrac{-k}{81}$
On expanding the equation we get,
$\dfrac{4}{3}+\dfrac{64}{27}=\dfrac{-k}{81}$
Take the L.C.M and add the terms, so we get,
$\dfrac{36+64}{27}=\dfrac{-k}{81}$
On further solving we get,
$\dfrac{100}{27}=\dfrac{-k}{81}$ $\Rightarrow 300=-k$
$\therefore k=-300$
Case 2: If $r=-\dfrac{4}{3}$ the equation will be,
$-\dfrac{4}{3}+{{\left( -\dfrac{4}{3} \right)}^{3}}=\dfrac{-k}{81}$
On expanding the equation we get,
$-\dfrac{4}{3}-\dfrac{64}{27}=\dfrac{-k}{81}$
Take the L.C.M and add the terms, so we get,
$\dfrac{-36-64}{27}=\dfrac{-k}{81}$
On further solving we get,
$\dfrac{-100}{27}=\dfrac{-k}{81}$ $\Rightarrow -300=-k$
$\therefore k=300$

So, the correct answer is “Option C”.

Note: The equation for sum of roots and product of roots must not be interchanged and we should be careful while using the signs. When the power of the variable is an even digit then the value can either be positive or negative so, both the values must be substituted to find the answer.