Question

If equation $p{x}^2+9x+3=0$ has real roots, then find value of $p$.

• A. $p\le \frac{-27}{4}$
• B. $p\le \frac{27}{4}$
• C. $p\le \frac{4}{27}$
• D. None

Answer 1

Answer: (B)

$p\le \frac{27}{4}$

Given equation is $p{x}^2+9x+3=0$

Here $a=p,b=9,c=3$

For real roots, $\Delta \ge 0$ i.e. $b^2-4ac\ge 0$

Therefore, $81-4\times p\times 3\ge 0$
$\Rightarrow 81-12p\ge 0$

$\Rightarrow 81\ge 12p$
$\Rightarrow p\le \frac{81}{12}$

$\Rightarrow p\le \frac{27}{4}$