Question

One root of the quadratic equations x2-bx+6=0 and x2-6x+c=0 is equal. The ratio of remaining roots is 3:4 If all the roots are positive integers find the values of b and c respectively

Answer 1

Step-by-step explanation:

Let roots of 1st equation BE A and B

And of second be A and C.

Also, it's given that, B/C = 3/4

Hence, 4B = 3C

We know,

Sum of roots = -b/a = b = A+B

product = c/a = 6 = AB

For second equation,

-b/a = 6 = A+C

c/a = c = AC

Hence, 6 = A+C = A+4B/3

18 = 3A+4B

18–3A/4 = B

Putting this in equation, AB = 6

(18–3A/4) A = 6

18A -3A² = 24

Rearranging,

=3A²-18A+24=0

Using quadratic formula,

18+- (324 - 288)½/ 6

18+-6/6

Hence, A = 4, 2

A= 2 (since the equation AB = 6 is not satisfied if A = 4, the result is not an integer as stated by the question.)