SOLUTION: one of the roots of the equation 5x^2 + kx = 4 is x = -2 the other root is x = ...what ? a -5/2 b -2/5 c 2/5 d 2 e 5/2

Question 252441: one of the roots of the equation 5x^2 + kx = 4 is x = -2 the other root is
x = ...what ?
a -5/2 b -2/5 c 2/5 d 2 e 5/2
Found 2 solutions by richwmiller, stanbon:
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
plug -2 in for x and solve for k
then solve the equation
5(-2)^2+k(-2)=4
5*4-2k=4
subtract 4 from each side
4*4-2k=0
add 2k to each side
16=2k
8=k
5x^2+8x=4
5x^2+8x-4=0

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=144 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 0.4, -2. Here's your graph:

.4=2/5
you could also have divided x+2 into the equation
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
one of the roots of the equation 5x^2 + kx = 4 is x = -2 the other root is
x = ...what ?
--------------------------
f(-2) = 0
5(-2)^2+k(-2)= 4
20-2k=4
2k = 16
k = 8
----------------
5x^2+8x-4 = 0
5x^2+10x-2x-4 = 0
5x(x+2)-2(x+2) = 0
(x-2)(5x-2) = 0
x = 2 or x = 2/5
------------------------
Ans: x = 2/5
============================
Cheers,
Stan H.

a -5/2 b -2/5 c 2/5 d 2 e 5/2

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