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Let $p,q,r,s$ be real numbers such that $pr=2(q+s).$ Consider the equations $x^{2}+px+q=0$ and $x^{2}+rx+s=0.$ Then

1. at least one of the equations has real roots.
2. both these equations have real roots.
3. neither of these equations have real roots.
4. given data is not sufficient to arrive at any conclusion.

recategorized | 23 views

+1 vote
$\underline{\mathbf{Answer: A}}$

$\underline{\mathbf{Explanation:}}$

The discriminant of the equation $\mathrm{x^2 + px + q = 0}$ is $\Delta_1 = \mathrm{p^2-4q}$

and discriminant of $\mathrm{x^2 = rx = s = 0}$ is given by $\Delta_2 = \mathrm{r^2 - 4s}$

Now, $\Delta_1 + \Delta_2 = \mathrm{p^2 - 4q + r^2 -4s = p^2 + r^2 -4(q+s) = p^2 + r^2 -2pr = (p-r)^2} \;[\text{if}\;\mathrm{pr = 2(q+s)}]$

$\therefore \Delta_1 + \Delta_2 = \mathrm{(p-r)^2} \ge 0$

$\because$ Sum of these two $\mathbf{Discriminant}$ is $\mathbf{positive}$.

$\therefore$ At least one of them has to be $\mathbf{positive}$.

$\therefore$ At least one of the equations between $\mathrm{x^2+px+q = 0}$ and $\mathrm{x^2 + rx + s = 0}$ must have real roots.

$\therefore \mathbf A$ is the correct option.
by Boss (18.9k points)
edited by
+1

Here needs some corrections. $\Delta_1+\Delta_2=(p-r)^2$ not $(p+r)^2$

$p^2+r^2-4(q+s)=p^2+r^2-2\{2(q+r)\}=p^2+r^2-2pr=(p-r)^2$

+1
Yes, thanks!