Solving for x

[elisemd]

I am not sure how to go about this problem. The square roots kinda confuse me...

Problem: Solve for x: 2x^2-4x = (1/16)x - 4
a. 2
b. 0,2
c. 4
d. 4 + 4 square root of 2
e. +4
f. -4
g. 0

I plugged in some of the numbers for x, which eliminated a few, but I still couldn't GET a right answer. I don't want to just guess. I want to figure it out. Please? Help?

~Elise[/list]

 

[stapel]

If you've plugged in all of the answers and can't get any of them to work, then either the correct answer ("none of the above") is not listed, or you're making a mis-step somewhere.

Please reply showing your work (at least a representative sample of two or three of the values being plugged in), so we can see what you're doing. Or please show what you did to try to solve the equation. (Since it's just a quadratic, did you, for instance, apply the Quadratic Formula?)

Thank you.

Eliz.

 

[elisemd]

My work on the "solving for x" problem

Original problem:

2x^2-4x = (1/16)x - 4

a. 2
b. 0,2
c. 4
d. 4 + 4 square root of 2
e. +4
f. -4
g. 0

a. 2 (2)^2 - 4(2) = (1/16) (2) -4
0 = (-31/8)
b. 2 (0) ^2 - 4 (0) = (1/16) (0) -4
0 = -4
c. 2 (4) ^2 - 4 (4) = (1/16) (4) -4
16= (-15/4)
d. I'm not sure how to put this one in.
e. Since positive 4 didn't work, then this can't be the correct answer right?
f. 2 (-4) ^2 - 4 (-4) = (1/16) (-4) -4
48 = 17/4
g. Since B didn't work with a 0 in it, this can't be the correct answer. But D isn't the correct answer...and I don't understand what I'm doing wrong.

 

[Gene]

What you typed looks valid (and agrees with your efforts) but none of the choices will work 'cause there is no real answer, only imaginary. Are you positive it is correct?

 

[stapel]

Thank you for replying showing what you'd tried. Here's the solution using the Quadratic Formula:


. . . . .\(\displaystyle \large{2x^2\,-\,4x\,=\,\frac{1}{16}x\,-\,4}\)


. . . . .\(\displaystyle \large{32x^2\,-\,64x\,=\,1x\,-\,64}\)


. . . . .\(\displaystyle \large{32x^2\,-\,65x\,+\,64\,=\,0}\)


. . . . .\(\displaystyle \large{x\,=\,\frac{-(-65)\,\pm\,\sqrt{(-65)^2\,-\,4(32)(64)}}{2(32)}}\)


. . . . .\(\displaystyle \large{x\,=\,\frac{65\,\pm\,\sqrt{4225\,-\,8192}}{64}}\)


. . . . .\(\displaystyle \large{x\,=\,\frac{65\,\pm\,\sqrt{-3967}}{64}}\)


Since we're left with a negative inside the square root, we can't possibly get any of the listed (real number) options as the solution.

Eliz.