Question

For what values of ‘p’ would the equation $x^2+2(p-1)x+p+5=0$ possess at least one positive root?

• A. [ , -5]
• B. [;, -1]
• C. [1, ]
• D. [2, ]

Answer 1

The correct option is A [ , -5]

For an equation to have positive roots D must be greater than '0'.

Now D from the equation $=4(p-1{)}^2-4(1\left)\right(p+5)=4p^2-12p-16$

$=4(p^2-3p-4)=4(p-4)(p+1)$

$\therefore 4(p-4)(p+1)>\mathrm{0......}\left(1\right)$

So, D is positive in the region $(-\infty ,-1)$ and $(4,\infty )$.

$x=\frac{-b+\sqrt{D}}{2a}\ge 0$
$⟹D\ge b^2$

$b^2=4(p-1{)}^2$
$\therefore p^2-3p-4\ge p^2+1-2p$
$⟹p\le -\mathrm{5..........}\left(2\right)$

Taking intersection of equation (1) and (2), we get -
p lies from $(-\infty ,-5]$.