Question

Find the values of $p$ and $q$ for which $x=\frac{3}{4}$ and $x=-2$ are the roots of the equation $p{x}^2+qx-6=0$

• A. $p=-4,q=5$
• B. $p=4,q=5$
• C. $p=4,q=-5$
• D. $p=-4,q=-5$

Answer 1

The correct option is B $p=4,q=5$

The quadratic equation is $p{x}^2+qx-6=0$ ......(i)

Given, $x=\frac{3}{4}$ and $x=-2$ are the roots of the quadratic equation.

Thus, putting $x=\frac{3}{4}$ in equation (i) we get,

$p{\left(\frac{3}{4}\right)}^2+q\times \frac{3}{4}-6=0$

$\Rightarrow p\times \frac{9}{16}+\frac{3q}{4}-6=0$

$\Rightarrow \frac{9p+12q-96}{16}=0$

$\Rightarrow 9p+12q-96=0$

$\Rightarrow 3p+4q-32=0$ ....(ii)
Now, putting $x=-2$ in equation (i), we get

$p(-2{)}^2+q(-2)-6=0$
$\Rightarrow 4p-2q-6=0$
$\Rightarrow 2p-q-3=0$ ....(iii)
Now, multiplying equation (iii) by $4$, we get
$\Rightarrow 8p-4q-12=0$ ....(iv)
Now, adding (ii) and (iv), we get
$11p-44=0$
$\Rightarrow p=\frac{44}{11}$

$\Rightarrow p=4$

Now, putting the value of p in equation (iii), we get
$\Rightarrow 2p-q-3=0$
$\Rightarrow 2\times 4-q-3=0$
$\Rightarrow 8-q-3=0$
$\Rightarrow -q=3-8$
$\Rightarrow -q=-5$
$\Rightarrow q=5$