Question

If x = −2 and x= −1/5 are solution of the equation 5x^2+ Px + R = 0 find the value of P
and R

Answer 1

Answer:

given x=-2 or -1/5

put both values of x in the above equation.

firstly x= -2

5(-2)^2+p(-2)+R=0

5*4-2p+R=0

20-2p+R = 0-------(1)

secondly x-1/5

5(-1/5)^2+p(-1/5)+R=0

5*(1/25)-p/5+R=0

1/5-p/5+R = 0------(2)

from (1) & (2), we have

20-2p+R = 1/5-p/5+R

20-2p+R = (1-p)/5+R

20-2p+R-R = (1-p)/5

20-2p = (1-p)/5

5(20-2p) = 1-p

100-10p = 1-p

100-1 = -p+10p

99 = 9p

p = 99/9

p=11

put the value of 'p' in the given equation

5(-2)^2+p(-2)+R = 0

5*4+11(-2)+R = 0

20-22+R = 0

-2+R = 0

R = 2