IIT JEE , AIEEE Forum - Mathematics , Algebra

srbhvatsa

Two roots of the equation x^3 +qx^2 +11x - p =0 are 2 and 3. Find p-q. (please solve in detail )

karthik2007

we are given the two roots. Since they are the roots of the equation, each of them must satisfy the equation.

Therefore, putting 2 in place of x, we get :

8 + 4q + 22 - p = 0 => 4q-p = -30

Also,
27 + 9q + 33 - p = 0 => 9q-p = -60

solving the two equations in p and q, we get :
q = -6, p = 10.

saarika

the procedure is right but on solving the equations

4q - p = -30

9q - p = -60

we get p = 6 nd q = -6

hence p - q = 6 - (-6) = 12

a5hw1n_5

let the third root be r

Therefore a=1

2+3+r=-b/a= -q ----------1 b=q

2*3+3*r+r*2= c/a=11--------2 c=11

2*3*r=-d/a= p----------3 d=p (the coeff of x^3,x^2,x, const

term respectively)

from 2 we get 5r=11-6

r=1

from 1 we get

q= -6

from 3 we get

p=6

therefore p-q= 12

karthik2007

I did the calculations in my mind.. so dont mind my mistakes.

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