Question

If one root is square of the other root of the equation $x^2+px+q=0$, then the relation between $p\text{and}q$ is

• A. $p^3–(3p–1)q+q^2=0$
• B. $p^3–q(3p+1)+q^2=0$
• C. $p^3+q(3p-1)+q^2=0$
• D. $p^3-q(3p-1)+q^2=0$

Answer 1

The correct option is A

$p^3–(3p–1)q+q^2=0$

Explanation for the correct option.

Let the one root of the given equation be $\alpha$ then the other root will be ${\alpha }^2$.

Now, we know that the sum of these roots will be equal to $-p$ and their product will be equal to $q$. From this we construct two equations,

$\alpha +{\alpha }^2=-p\dots \left(1\right)$

${\alpha }^3=q\dots \left(2\right)$

Now, in equation $\left(1\right)$, apply the cube on both the sides,

$$\begin{gathered} \therefore {\left({\alpha }^2+\alpha \right)}^3={\left(-p\right)}^3 \\ \Rightarrow {\left\{\alpha \left(\alpha +1\right)\right\}}^3=-p^3 \end{gathered}$$

Apply the algebraic identity ${\left(a+b\right)}^{\mathbf{3}}\mathbf{=}{\mathit{a}}^{\mathbf{3}}\mathbf{+}{\mathit{b}}^{\mathbf{3}}\mathbf{+}\mathbf{3}\mathit{a}\mathit{b}\left(a+b\right)$, then

$\begin{array}{rcl}& \Rightarrow & -p^3=a^3\left(a^3+1+3a\left(a+1\right)\right)\end{array}$

Apply the value of $a^3$ from equation $\left(2\right)$,

$\begin{array}{rcl}\therefore -p^3& =& q\left(q+1+3\left(-p\right)\right)\\ & \Rightarrow & p^3+q^2+q-3pq=0\\ & \Rightarrow & p^3-\left(3p-1\right)q+q^2=0\end{array}$

Hence, the correct option is (A).