Question

Two quadratic equations $x^{2}$ - bx + 6 and $x^{2}$ - 6x + c have a common root. If the remaining roots of the first and the second equations are positive integers and are in the ratio 3:4 respectively, then the common root is?
A) 1
B) 2
C) 3
D) 4

Answer 1

Answer:

D) 4 is correct

Step-by-step explanation:

x² - bx + 6=0----------(1) and

x²- 6x + c=0-------------(2)

let common root is p

then

p² - bp + 6=0 and

p²- 6p + c=0

on subtracting we get

p(6-b)+6-c=---------..(3)

let the other roots are 3q and 4q

then roots  of eqn (1) are p and 3q

so p*3q=6,

q=2/p

roots  of eqn (2) are p and 4q

so p+4q=6

p+4*2/p=6

p²-6p+8=0

(p-2)(p-4)=0

p=2 or 4

p=4

D) 4 is correct

The question is very interesting