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 $x^2-15x+56=0$. The sum of all the four roots is:

• A. $17$
• B. $18$
• C. $19$
• D. $23$

Answer 1

The correct option is D $23$

Let the solution of first equation be $a,b$

and for 2nd equation be $a,c$

Given, $ab+ac+a^2+ba+bc+ac=192$

$2a(b+c)+a^2+bc=192$ ..... (1)

Now, $x^2-15x+56=0$ has uncommon roots, i.e. $b,c$

So, in eq (1) we can substitute $b+c=15,bc=56$ and obtain,

$2a\left(15\right)+a^2+56=192$

$\Rightarrow a=4$ (positive root)

So, sum of all $4$ roots is

$2a+(b+c)=8+\left(15\right)=23$