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

Answer:

#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#