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 sum of all the four roots.

Answer 1

x2 -15x +56 =(x-7)(x-8)

Answer 2

Then?

Answer 3

i think the sum of all 4 roots is 30.6..

Answer 4

let a, b, roots of the 1st eq.
a, c roots of the 2nd eq.

The equation whose roots are the two different roots is x^2 -15x +56=0
(x-b)(x-c) = x^2 - 15x +56
= (x-8) (x-7)
so,
b=8 and c=7

The sum of the product of all four roots taken two at a time is 192 so,
ab + ac = 192 /(from th given values of b and c)
8a + 7a = 192
15a = 192
a = 12.8

the sum of all 4 roots is
a+a+b+c = 2(12.8) +8 +7
= 30.6

Answer 5

I get 23.

Answer 6

Let the four roots be\[\alpha, \beta, \gamma, \delta>0\]where the first two roots belong to the first quadratic and the last two to the second. Without loss of generality, let the two roots that are shared be alpha and gamma. Then,\[\alpha = \gamma\]We're told the sum of the product of the roots taken two-at-a-time is 192; that is,\[\alpha \beta + \alpha \gamma + \alpha \delta +\beta \gamma + \beta \delta + \gamma \delta=192 \rightarrow 2 \alpha \beta + 2 \alpha \delta + \beta \delta + \alpha^{2}=192\]Now, we're also told that the remaining two roots, beta and delta, are also roots of the quadratic,\[x^{2}-15x+56=0\]The roots of this quadratic are\[x=\frac{15 {\pm} 1}{2} \rightarrow \beta = 8, \delta = 7\](doesn't matter whether beta = 8 and delta = 7 or vice versa). Substitute this into the '192' equation above,\[16 \alpha + 14 \alpha + 56 + \alpha^{2} = 192 \rightarrow \alpha^{2}+30 \alpha -136=0 \rightarrow \alpha = 4,-34\]Since we're told all roots are positive, we must take \[\alpha = 4\]The sum of the roots is therefore,\[\alpha + \beta + \alpha + \delta = 4 + 8 + 4 + 7 = 23\]