Find ‘K’ If The Sum Of The Zeros Of The Polyomial X (Power 2) (K+6)X + 2(2k1) Is Half Their Product

closed

Answers:

If  p and q are the roots of the equation     x² +bx+c then:

p+q=(-b)/a   =   -(-(k+6))/1

p*q=c/a       =    2(2k-1)/1

According to the information:  p+q=1/2(p*q)

k+6=(1/2) *(2*( 2k-1))

solve it and see that it  comes to…..

k+5=2k-1    [subtract both sides by k]

5=k-1

k=5

SOLVED,

HOPE YOU UNDERSTOOD,GOOD LUCK!!!