Question

Two quadratic equations with positive roots have one common root. The sum of the product of all four roots taken two at a time is $192$. The equation whose roots are the two different roots is $x^2-15x+56=0.$ Find the common root.

• A. $-34$
• B. $18$
• C. $-32$
• D. $4$

Answer 1

The correct option is D $4$

Given that, two quadratic equations with positive roots have common root

Let a,b be roots of one equation and b,c are roots of another equation with b as common root.

Sum of product of roots taken two at a time $=192$

$b^2+ac+2ab+2bc=192$

$b^2+ac+2b(a+c)=192\to 2$

$x^2-15x+56=0$

$(x-7)(x-8)=0$

Roots of $x^2-15x+56=0$ are roots other than common root

$\therefore a=7,c=8$ (or ) $a=8,c=7$

$a+c=15,ac=56$

$\Rightarrow b^2+56+30b=192$

$b^2+30b-136=0$

$b^2+34b-4b-136=0$

$(b-4)(b+34)=0$

$\therefore b=4$ ($\because b\ne -34$ as $b>0$)

$\therefore$Common root is $4$