Find k if sum the zeros of x²-(k+6)x+ 2(k-1) is half of the product
Answers
Answered by
5
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,
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,
Similar questions