Question

If both roots of the quadratic equation $x^2+4px+6p^2+3p-2=0$ are less than $4$, then $p$ lies in the interval

• A. $(-2,\frac{1}{2}]$
• B. $[-2,\frac{1}{2}]$
• C. $(-2,\frac{-7}{6})$
• D. $(\frac{-7}{6},\frac{1}{2}]$

Answer 1

The correct option is D $(\frac{-7}{6},\frac{1}{2}]$

Given $x^2+4px+6p^2+3p-2=0$
Now $D\ge 0$
$16p^2\ge 4\left(1\right)(6p^2+3p-2)$
$8p^2+12p-8\le 0$
$(8p-4)(p+2)\le 0$
$p\in [-2,\frac{1}{2}]...\left(1\right)$
Also $f\left(4\right)>0$
$\Rightarrow 6p^2+19p+14>0$
$p\in (-\infty ,-2)\cup (\frac{-7}{6},\infty )...\left(2\right)$
Also $\frac{\text{sum of roots}}{2}<4$
$\frac{-4p}{2}<4$
$p>-2$
so from $\left(1\right),\left(2\right)$ and $\left(3\right)$
we get $p\in (\frac{-7}{6},\frac{1}{2}]$