Show that if $p, q, r, s$ $p,q,r,s$ are real number and $p r = 2 (q + s)$ $pr=2(q+s)$ then at-least one of the equations $x^{2} + p x + q = 0$ $x^2+px+q=0$ and $x^{2} + r x + s = 0$ $x^2+rx+s=0$ has real roots?

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Answer 1 · 2018-01-08T08:55:42

Please see below.

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

and that of $x^2+rx+s=0$ is $Delta_2=r^2-4s$

and $Delta_1+Delta_2=p^2-4q+r^2-4s$

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

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

= $(p+r)^2-2[pr-2(q+s)]$

and if $pr=2(q+s)$ , we have $Delta_1+Delta_2=(p+r)^2$

As sum of the two discriminants is positive,

at least one of them would be positive

and hence atleast one of the equations $x^2+px+q=0$ and $x^2+rx+s=0$ has real roots.

Show that if p,q,r,sp,q,r,sp,q,r,s are real number and pr=2(q+s)pr=2(q+s)pr=2(q+s) then at-least one of the equations x^2+px+q=0x2+px+q=0x^2+px+q=0 and x^2+rx+s=0x2+rx+s=0x^2+rx+s=0 has real roots?