ISI2016-DCG-30

Question

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

• A. at least one of the equations has real roots.
• B. both these equations have real roots.
• C. neither of these equations have real roots.
• D. given data is not sufficient to arrive at any conclusion.

Answer 1

$\underline{\mathbf{Answer: A}}$

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

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.

Comments

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

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

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Yes, thanks!