Question

If $3p^2=5p+2$ and $3q^2=5q+2$, where $p\ne q$, then the equation, whose roots are $3p-2q$ and $3q-2p$, is

• A. $3x^2–5x–100=0$
• B. $5x^2+3x+100=0$
• C. $3x^2–5x+100=0$
• D. $5x^2–3x–100=0$

Answer 1

The correct option is A

$3x^2–5x–100=0$

Explanation for the correct option:

Step 1. Find the equation:

Given, $3p^2=5p+2$

$\Rightarrow$ $3p^2–5p–2=0$

$\Rightarrow$$(3p+1)(p–2)=0$

$\Rightarrow$ $p=-\frac{1}{3},2$

Similarly, $3q^2=5q+2$

$\Rightarrow$$3q^2–5q–2=0$

$\Rightarrow$ $q=-\frac{1}{3},2$

It’s given that $p\ne q$

The roots are $(3p–2q)$ and $(3q–2p)$.

Step 2. Put the value of $p$ and $q$ in the roots:

$\begin{array}{rcl}\alpha & =& 3p–2q\\ & =& 3\times 2–2\times -\frac{1}{3}\\ & =& 6+\frac{2}{3}\\ & =& \frac{20}{3}\end{array}$

$\begin{array}{rcl}\beta & =& 3q–2p\\ & =& 3\times -\frac{1}{3}–2\times 2\\ & =& -5\end{array}$

$\begin{array}{rcl}\alpha +\beta & =& \frac{20}{3}+(-5)\\ & =& \frac{5}{3}\end{array}$

$\begin{array}{rcl}\alpha \beta & =& \frac{20}{3}\times (-5)\\ & =& –\frac{100}{3}\end{array}$

Step 3. The required equation is $x^2–(\alpha +\beta )x+\alpha \beta =0$

$\Rightarrow$ $x^2–\frac{5}{3}x–\frac{100}{3}=0$

$\Rightarrow$ $\mathbf{3}{\mathit{x}}^{\mathbf{2}}\mathbf{}\mathbf{–}\mathbf{}\mathbf{5}\mathit{x}\mathbf{}\mathbf{–}\mathbf{}\mathbf{100}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$

Hence, Option ‘A’ is Correct.