Solve for x?

10x² + 3x³ = 8x + 0^x
10x² + 3x³ = 8x + 0
3x³ + 10x² - 8x = 0
x(3x² + 10x - 8) = 0
x(3x-2)(x+4) = 0

x = 0 or 2/3 or -4 but for 0^x, 0^0 and 0^-4 is undefined.

Therefore, x = 2/3

2006-08-31 23:17:14 · answer #1 · answered by Anonymous · 5⤊ 1⤋

First, state the domain, which is all the permissible values for x.
Since in 0^x, x can't be equal to 0 or a negative number, the domain is: {x|x Îµ R, x>0}.
Now, to solve.
10xÂ² + 3xÂ³ = 8x + 0
10xÂ² + 3xÂ³ - 8x = 0
3xÂ³ + 10xÂ² - 8x = 0
x(3xÂ² + 10x - 8) = 0
x(3xÂ² + 12x - 2x - 8) = 0
x[(3xÂ²+12x) - (2x+8)]=0
x[3x(x+4) -2(x+4)]=0
x(x+4)(3x-2)=0
Solving for the value of x, we get x=0, -4, 2/3.
But since 0 and -4 are not part of the domain, the only possible value for x is x=2/3.
SS: {x|x = 2/3}

2006-09-02 05:42:14 · answer #2 · answered by fictitiousness ;-) 2 · 0⤊ 0⤋

... + 0^x is a little strange -- it doesn't do anything except saying that x can't be zero

The rest of the equation simplifies as

10 x^2 + 3 x^3 - 8 x = 0

(3 x^2 + 10 x - 8) x = 0

this gives the solution x = 0, but since 0^0 is not possible it is illegal

3 x^2 + 10 x - 8 = 0

Is a factoring possible? If so, the 3 must arise as 1 * 3

(x + ...) (3 x + ...) = 3 x^2 + 10 x - 8 = 0

(x + 4) (3 x - 2) = 0

x + 4 = 0 ... or ... 3 x - 2 = 0

x = -4 ... or ... x = 2/3

2006-09-01 09:02:53 · answer #3 · answered by dutch_prof 4 · 0⤊ 0⤋

The importand point to consider is 0^0. What is its value? Many say its indeterminable. But I found an article from experts in MathForum that 0^0 = 1.

Please check the link to see why it is so

If that is the case, we are faced with problems.

let us take the case x not equals 0

This makes 0^x = 0

when we solve this case x=2/3 , -4 or 0. That is when we take a case x not equals zero and solve, we get x=0!!!

And if we substitue x=0 in the given equation itself, we get

0 + 0 = 0 + 1, according to what the experts say.

This is because the experts have applied rules like L'Hospitalz rule to arrive at the solution.

Also for the case of x=-4 , we land up with similar problems. Therefore x=2/3 is the ONLY DEFINITE SOLUTION..

2006-09-01 06:18:43 · answer #4 · answered by Truth Seeker 3 · 4⤊ 0⤋

10x^2 + 3x^3 = 8x + 0^x
10x^2 + 3x^3 = 8x +0
x(3x^2 +10x -8)=0
x (3x-2 )(x+4 )=0
Therefore x=0,2/3, or-4

2006-09-01 10:39:00 · answer #5 · answered by elitetrooper459 3 · 0⤊ 0⤋

Depending on your definition of 0^0, 0 might be an answer... but usually 0^0=1, in which case x=0 is not valid.

The other answers are x=-4 and x=2/3, but 0^-4 is another problem:
0^-4 =
1/(0^4) =
1/0

and 1/0 is pretty much undefined.

So the only valid answer is 2/3

2006-09-01 06:21:25 · answer #6 · answered by robcraine 4 · 2⤊ 0⤋

x=2/3

2006-09-01 08:17:54 · answer #7 · answered by bkbarile 5 · 0⤊ 0⤋

Answers: 0, -4, 2/3

solution:

3x^3 +10x^2 - 8x = 0
x(3x2 +10x -8) = 0
x(3x - 2)(x + 4) = 0

3x = 2 -->x = 2/3
x +4 = 0 --> x = -4

2006-09-01 06:19:27 · answer #8 · answered by alimoalem2000 2 · 1⤊ 2⤋

x=0 (or) x=-4 (or) x=2/3
But, if x=0, then,
0^x=0^0 which is not determined
and if x=-4, then,
0^x=0^-4=1/0^4=1/0 which can also not be determined

As such, the only possible solution is x=2/3

Have a good day!

2006-09-01 07:37:38 · answer #9 · answered by Bhushan S 1 · 1⤊ 0⤋

x= 2/3
x= -4

2006-09-01 08:49:48 · answer #10 · answered by Anonymous · 0⤊ 0⤋

15 answers