PLZ HELP???????? URGENT...?

9x^2-27x+k = 0

now remember that summation of roots is equal to -b/a = 27/9 = 3

thus alpha+beta=3 (1)

on the other hand, multiplication of roots is equal to c/a = k/9

thus: 9*alpha*beta=k (2)

now if you want 2*alpha+5*beta = 7 you only need to replace alpha and beta from (1) and (2) as a function of k that is

alpha=3-beta and (3-beta)*9*beta = k

==> (27-k)/9=beta^2 ==> beta = 1/3*(27-k)^.5

2alpha+5beta = (2alpha+2beta)+3beta = 2*3 + (27-k)^.5=7

==> 27-k = 1 ==> k = 26

I somewhere made a mistake. The correct answer is K=8 as mentioned by others

Here

9x^2-27x+k = 0

now remember that summation of roots is equal to -b/a = 27/9 = 3

alpha+beta=3 (1)

on the other hand, multiplication of roots is equal to c/a = k/9

thus: 9*alpha*beta=k (2)

2*alpha+5*beta = 7(3) you only need to replace alpha and beta from (1) and (2) as a function of k that is

alpha=3-beta and put it in equation 3..you will get alpha =8/3 and beta=1/3 ..now k =9 *alpha*beta =8

9x² - 27x + k = 0

(3x - 9/2)² = 81/4 - k

3x - 9/2 = ±√(81/4 - k)

3x = 9/2 ± √(81/4 - k)

x = 3/2 ± √(9/4 - k/9)

2(3/2 + √(9/4 - k/9)) + 5(3/2 - √(9/4 - k/9)) = 7

=> 3 + 15/2 - 3√(9/4 - k/9) = 7

=> 7/6 = √(9/4 - k/9)

=> 49/36 = 9/4 - k/9 = (81 - 4k)/36

=> k = 8

or

9x² - 27x + k = 9(x - α)(x - β) = 9(x² - (α + β)x + αβ)

α + β = 3

2α + 5β = 7

=> β = 1/3, α = 8/3

=> k = 9αβ = 8

1st of all we have to put the problem in equation form, lets take alpha to be A and beta to be B then (x plus A)(x plus B)=Xsquared -3x plus k/9=0 again by comparing the above equation. Therefore -(A plus B)=-3...equ1 and AB=k/9...equ3 from the last statement in the question 2A plus 5B=7...equ2 then by solving equ1 and equ2 simultaneously we get that A=8/3, B=1/3 finally substitute A and B into equ3 we get that k=8 thanks