I get A

2x^2 + 3x – k = 0 ---> dividing by 2 ---> x^2 + 3/2x - k/2 = 0

this can be factorize as (x+a)(x+b) where

a * b = -k/2
a + b = 3/2

we know that one of the solutions is x = 6 so a = -6 or b = -6

a + b = 3/2 --> -6 + b = 3/2 ---> b = 15/2

-6 * 15/2 = - K/2 ---> 6 * 15 = K ---> K = 90

What it means for 6 to be a root of f(x) = 2x^2 + 3x - k is that f(6) = 0.

So 2*6^2 +3*6 - k = 0
72 + 18 - k = 0
k = 90

The answer is A. Where did you get this problem?

One way I found useful is that k must be divisible by 6, so that eliminates D and E.

I wrote the factored form out so it would be better so see:

(2x + ?)(x - 6) = 0

x has to be in the (x-6) because one of the values for x = 6 so -6 and +6 would add up to produce zero.

From there we can see that 15 fits in perfectly because it gives us the +3x value.

oks wrote:What's a quick method to solve the problem below (other than plugging in)?

7. If one root of the equation 2x2 + 3x – k = 0 is 6, what is the value of k?
(A) 90
(B) 42
(C) 18
(D) 10
(E) –10

OA is E

Test for which of the answer choices this is a perfect square

(b^2 -4ac)^.5=(9+8k)^.5 =(729)^.5=27. First test reveals answer
Choose A. Handy to know first 30 roots.

We can use the method called completing the square

(X+3/4)^2 = (8K+9)/4^2

when X = 6 .........27^2 = 8K+9

8K = 30*24

K = 90

I got this from the Plus series tests. My answer was also A - which is 90. I guess it's an error in the Plus test answers.