Help please?

Question

Okay, I'm having A TON of trouble with this problem, PLEASE help! The quadratic formula: Solve the quadratic equation by using the quadratic formula. If the
roots are irrational, give the rational approximations to the nearest tenth.
Please, FILL in the blanks!

x^2/3 + 1/2 = (5/6)x

{___,___}

Thanks!

Answer 1

To use the quaratic formula, you must have the equation set equal to zero, so subtract (5/6)x from both sides.

(1/3)x² + (-5/6)x + (1/2) = 0.

From here, substituting into the formula using a = 1/3, b = -5/6, and c = 1/2 is doable, but ugly. I'd multiply both sides by a common denominator of 6 to get rid of all your fractions, first.

(1/3)x² + (-5/6)x + (1/2) = 0.
6[(1/3)x² + (-5/6)x + (1/2)] = 6[0].
2x² - 5x + 3 = 0.

I'd factor this, too, but you said use the quadratic formula, so...
x = {-(-5) ± √[(-5)² - 4(2)(3)] } / [2(2)]
x = {5 ± √[25 - 24] } / 4
x = {5 ± 1 } / 4
x = 3/2 or x = 1.
(Any time you get rational roots to a quadratic, the expression is factorable, so there's no need to go through all this junk!)

If you're bent on using the fractions, it's trickier, but you'll get the same solutions:

x = {-(-5/6) ± √[(-5/6)² - 4(1/3)(1/2)] } / [2(1/3)]
x = {5/6 ± √[(25/36) - (2/3)] } / (2/3)
x = {5/6 ± √[(25/36) - (24/36)] } · (3/2)
x = {5/6 ± √(1/36) } · (3/2)
x = {5/6 ± 1/6 } · (3/2)
x = {5/4 ± 1/4 }
x = 3/2 or x = 1

Answer 2

Is the question (x^2)/3 +1/2 = (5/6)x? or x^(2/3) + 1/2 = (5/6)x?

Answer 3

Check a site for the quadratic formula.

Answer 4

(x^2)/3 + 1/2 = (5/6)x
(1/3)x^2 - (5/6)x + (1/2) = 0

Multiply everything by 6:
2x^2 - 5x + 3 = 0
(2x-3)(x-1) = 0
2x-3 = 0 or x-1 = 0
x = 3/2 or 1
x = 1.5 or 1

Answer 5

multiply each term by 6

2x^2-5x+3=0

a=2
b=-5
c=3

x1=-3/2

x2=-1

Answer 6

holy crap. uhh, :()