Question

two quadratic equations with positive roots have one common root. Thea 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 different roots is

Answer 1

Two quadratic equations with positive roots have one common root.

Let 'a' and 'b' be the roots of first quadratic equation and 'a' and 'c' be the roots of second quadratic equation.

Since, the sum of the product of all four roots taken two at a time is 192.

$(a \times b) +( c \times a)+( b \times c)+(a \times a) = 192$

$ab + ca + bc + a^2 = 192$    (Equation 1)

Now, the equation whose roots are the two different roots is

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

By splitting the middle term, we get

$x^2 -7x - 8x + 56=0$

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

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

Therefore, x = 7 and x =8

Therefore, the two different roots of one equation are 7 and 8.

Let a = 7 and b = 8

Substituting the values in equation 1 , we get as

$(7 \times 8) + ( c \times 7)+(8 \times c)+( 7 \times 7) = 192$

$56+7c+8c+49=192$

$105+15c = 192$

$15c = 192-105 = 87$

$c =\frac{87}{15}$

c = 5.8

So, the sum of all different roots = a+b+c

= 7+8+5.8

= 20.8

Therefore, the sum of all different roots is 20.8