Question

lf $pr=2(q+s)$, then among the equations $x^2+px+q=0$ and $x^2+rx+s=0$

• A. both have real roots
• B. both have imaginary roots
• C. at least one has real roots
• D. at least one has imaginary roots

Answer 1

Answer: (C)

at least one has real roots

Given $pr=2(q+s),x^2+px+r=0,x^2+rx+s=0$

Let $D_1$, and $D_2$ be discriminants of

$x^2+px+r=0$ and $x^2+rx+s=0$

$then{D}_1=p^2-4q;D_2=r^2-4s$

$D_1+D_2=p^2+r^2-4(q+s)$

$=p^2+r^2-2pr$

$=(p-r{)}^2\ge 0.$

$\therefore D_1+D_2\ge 0$

$\Rightarrow$ Atleast one among $D_1$ and $D_2$ is greater than zero .

$\Rightarrow$ Equation has atleast one has real root .