If the roots of equation x^2 + qx + p=0 are each 1 less than the roots of the equation x^2 + px + q = 0 then (p+q) is equal to?

2 Answers
Apr 1, 2018

given that the roots of equation x^2 + qx + p=0 are each 1 less than the roots of the equation x^2 + px + q = 0.

Let the roots of the equation x^2 + px + q = 0 be alpha and beta . So the roots of equation x^2 + qx + p=0 will be (alpha-1) and (beta-1)

So we can write

alpha+beta=-pand alphabeta=q

again

alpha+beta-2=-qand (alpha-1)(beta-1)=p

Comparing we get

alpha+beta-2=-alphabeta.........[1]

and

(alpha-1)(beta-1)=-alpha-beta

=>alphabeta-alpha-beta+1=-alpha-beta

=>alphabeta=-1 .....[2]

By {1] and [2]

alpha+beta-2=-(-1)=1

alpha +beta=3

Now from first two relations

q-(-p)=alphabeta-(alpha+beta)=-1-3=-4

=>p+q=-4

Apr 1, 2018

p+q=-4

Explanation:

Assuming

x^2+qx+p = (x-x_1)(x-x_2) = 0
x^2+px+q = (x-x_1+1)(x-x_2+1) = 0

and comparing coefficients we get

{(p=-(x_1+x_2)),(q = x_1 x_2),(p = 1-(x_1+x_2)+x_1x_2),(q = 2-(x_1+x_2)):}

now calling y_1 = x_1+x_2 and y_2 = x_1x_2 we have

{(p=-y_1),(q = y_2),(p = 1-y_1+y_2),(q = 2-y_1):}

Solving

p=-3,q=-1, y_1=3, y_2 = -1

hence

p+q = -4