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# Can someone please help with math?

I have a test tomorrow which basically decides if I go to med-school or not. I am having trouble with an simple algebraic function.

-2(x-3)^3 + 16 = 0 (solve for x)

I can do some steps

-2(x-3)^3 = -16

(x-3)^3= -8

I just don't know how to take the cube from both sides. If I don't learn this within the next few hours, I am in a lot of trouble.

I appreciate any help in advance!!!!

Thank you all!! You guys have all been really helpful, definately saved my skin! The weird thing is that I can do all the calculus, statistics and probability, but I forgot basic algebra 0:

Again, I cannot thank you guys enough!

### 9 Answers

- PuggyLv 71 decade agoFavorite Answer
-2(x - 3)^3 + 16 = 0

I'm not sure if you want complex roots too, but in case you do, first divide by (-2).

(x - 3)^3 - 8 = 0

Now factor as a difference of cubes. A reminder that

a^3 - b^3 = (a - b)(a^2 + ab + b^2)

[(x - 3) - 2] [ (x - 3)^2 + 2(x - 3) + 4 ] = 0

Simplify.

[x - 5] [ x^2 - 6x + 9 + 2x - 6 + 4 ] = 0

[x - 5] [ x^2 - 4x + 7 ] = 0

At this point, you equate each factor to 0 and solve individually.

x - 5 = 0

x^2 - 4x + 7 = 0

1) x - 5 = 0

x = 5

2) x^2 - 4x + 7 = 0

Use the quadratic formula.

x = [-b ± √(b² - 4ac)]/(2a)

x = [4 ± √(16 - 4(7))]/(2)

x = [4 ± √(16 - 28)]/(2)

x = [4 ± √(-12)]/(2)

x = [4 ± √(4 * 3 * (-1))] / 2

x = [4 ± 2i√(3)]/2

x = 2 ± i√(3)

Therefore, all of your solutions are

x = { 5, 2 + i√(3), 2 - i√(3) }

Again, I don't know how advanced the math you're taking is; it may not even want the complex solutions. But there they are, just in case.

Good luck in med school.

- Anonymous1 decade ago
The first step you did is right. The second one isn't: (x-3)^3=8 and not -8. Finding the cube of 8 means which number times itself times itself is 8. Since the 8 is positive, the number you need also has to be a positive number. So, the number you are after is 2 since 2 times 2 is 4 and 4 times 2 is 8. Hence x-3=2 or x=5.

- 1 decade ago
Next step, factor -8 into prime factors (-2*-2*-2). Take the cube root, by pulling out all 3 of the same number, and writing only one two (it's negative). You now have (x-3)=-2. Add 3 to both sides, you have x=1.

- detektibgapoLv 51 decade ago
-2x(x-3)^3 + 16 = 0

-2(x-3)^3 = -16

(x-3)^3 = 8 (not -8 because you are dividing two negative numbers ( -16/-2 = 8)

Get the cube root of both sides

x - 3 = 2

x = 5

To check

-2(5-3)^3 + 16 = 0

-2(2)^3 + 16 = 0

-2(8) + 16 = 0

-16 + 16 = 0

0 = 0

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- 1 decade ago
yeah! you were on the right track initially

but in your 2nd step, when you divide both sides by -2, you should have

(x-3)^3 = 8

since 2 is the cube root of 8,

(x-3)^3 = 2^3

x-3 = 2

x = 3+2

x = 5

- durrettLv 44 years ago
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- 1 decade ago
I will start from your last step:

-(x-3)^3= -8 => (you had a small mistake here)

(x-3)^3 = (2)^3 ((2)^3 =8)

=> x-3 = 2 => x=5

- gudspelingLv 71 decade ago
(x-3)^3 = 8

Remember that 8=2^3

(x-3)^3=(2)^3

Take cube roots of both sides of the equation

x-3=2

x=2+3=5

x=5

- 1 decade ago
I don't think anyone should help you, nobody wants a doctor that can't figure this stuff out.

How are you going to calculate how much medicine a person requires? Don't you need to come up with functions based on a person's health status using many many parameters? This is a simple expression with one unknown.