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, ]
• E. none of these.

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 $=4(p-1{)}^2-4(1\left)\right(p+5)=4p^2-12p-16$

$=4(p^2-3p-4)=4(p-4)(p+1)......\left(i\right)$
$x=[-b+\sqrt{D}]\ge 0$
$D\ge b^2$
$b=4(p-1{)}^2$
$p^2-3p-4\ge p^2+1-2p$
$p\le -\mathrm{5..........}\left(2\right)$
Taking intersection of equation 1 and 2
$pE[-\infty ,-5]$

So roots for eq. (i) are : -1 and 4 and comparing with $a{x}^2+bx+c$ , we have 'a' $as>0$ so function will be open ended in the upward direction.

So, , D is positive in region $[-\infty ,-1]$ and $[4,\infty ]$. So for equation to posses at least one positive 'p' should either lie in $[-\infty ,-1]$ or in $[4,\infty ]$. So answer is option (b).