Running into trouble with a couple problems...
#1)
√(8x+1) = 2x-1
I keep getting 4x^2-12x+0 when writing in standard form, need to solve for x and know the answer is 3. This seems wrong that c=0. Where am I messing up?
----
#2)
Also solve.. ^3√(5x-2)=2
2007-05-08 06:53:00 · 15 answers · asked by Randomblina >^..^< 3 in Science & Mathematics ➔ Mathematics
√(8x+1)=2x-1
8x+1=4x^2-4x+1
4x^2-12x=0
4x(x-3)=0
hence either 4x=0.........or........x=3(u were right u just needed to take x common)
2.cube on b.s
5x-2=8
5x=10
x=2
2007-05-08 06:59:42 · answer #1 · answered by pjd_0501 2 · 0⤊ 1⤋
1) square both sides and you get 8x+1=4x^2-4x+1
put everything on the same side and you get what you got 4x^2-12x=0
Factor out a 4x and you get 4x(x-3)=0
You know that either 4x or x-3 is equal 0 so you know that the two possible solutions are x=0 so x=3.
The answer cannot be 0 because if you plug a zero in, you get 1=-1, which cannot be true, so your answer is 3.
2. Cube both sides and you get 5x-2=8
solve for x and you get 2
2007-05-08 07:03:42 · answer #2 · answered by Soccer Tease 4 · 1⤊ 0⤋
Take the square of both sides, and that gets rid of the radical.
8x + 1 = (2x - 1)^2
8x + 1 = 4x^2 - 4x + 1
8x = 4x^2 - 4x
4x^2 - 12x = 0.
Why do you think it's "wrong" that c=0? It can happen. From here, you can just factor out an x. Or even better, a 4x:
4x(x - 3) = 0
So x = 0 or 3
For the second problem, I take it "^3√" means the cube root. In this case, cube both sides to get rid of the radical:
5x-2 = 2 ^3
5x - 2 = 8
5x = 10
x = 2
2007-05-08 07:01:29 · answer #3 · answered by Anonymous · 0⤊ 1⤋
First x>=-1/8 and 2x-1>=0 so x>=1/2Now squaring both sides
8x+1=4x^2-4x+1
so4 x^2-12x =0 4x(x-3) = 0 x=0 Not a solution and x=3
I suppose you meant cubicroot(5x-2)=2
5x-2=2^3 so x=2
2007-05-08 07:06:21 · answer #4 · answered by santmann2002 7 · 1⤊ 0⤋
Squaring both sides yields:
8x+1=4x^2-4x+1
-4x^2+12x=0
x(-4x+12)=0
The values of x that satisfy this equation are x=0 and x=3
2007-05-08 06:58:32 · answer #5 · answered by helper 7 · 0⤊ 1⤋
Start by squaring each side, then solving it.
8x + 1 = 4x^2 - 4x + 1
0 = 4x^2 - 12x
0 = 4x(x - 3)
0 = 4x
0 = x
0 = x - 3
3 = x
For #2, raise both sides to the third power and then solve:
5x - 2 = 8
5x = 10
x = 2
2007-05-08 07:02:01 · answer #6 · answered by Jason P 4 · 0⤊ 1⤋
1.
first square both sides to get ride of the radical...
so..
8x+1 = 4x^2 -4x+1
then move the 8x+1 to the other side to set the equation to 0
so..
4x^2-12x = 0
then you can GCF the equation
so...
4x(x-3)=0
then set both of those with x to 0
so...
4x=0 x=0
and
x-3=0 x=3
0 and 3 are both answers... you didn't have the equation wrong at all you just needed to GCF the equation then solve...
2.
cube both sides
so..
5x-2 = 8
subtract 2
5x=10
divide by 5
x=10
2007-05-08 07:00:51 · answer #7 · answered by Anonymous · 0⤊ 2⤋
helper - I get your math, but I think I made a wrong turn somewhere.
If x = 0 is a solution
then
√(8*0 +1) = 2*0 -1
√(1) = -1
1 = -1
Any idea where I went wrong? Or is this because the original equation involved a square root and thus only really has 1 solution?
2007-05-08 07:08:23 · answer #8 · answered by Anonymous · 1⤊ 0⤋
x=3 is the only solution to this equation. x=0 is not a solution to the original equation because plugging it in gives 1=-1.
When you square* both sides of an equation, you have to check your answers because the equation that you get can have more solutions than the original. In this case we get 8x+1=4x^2-4x+1, which has two solutions, only one of which is a solution to the original.
*or raise both sides to any even power.
2007-05-08 07:11:20 · answer #9 · answered by TurkeyGobbler 2 · 0⤊ 0⤋
#1 is fine, 4x² - 12x = 0 → 4x(x-3) = 0
→ x=0 or x=3.
2007-05-08 07:00:48 · answer #10 · answered by Anonymous · 0⤊ 0⤋
Question 1
8x + 1 = 4x² - 4x + 1
4x² - 12x = 0
4x.(x - 3) = 0
x = 0, x = 3
Question 2
8x + 1 = 4x² - 4x + 1
4x² - 12x = 0
4x.(x - 3) = 0
x = 0, x = 3
Question 3
Not sure what you mean. Could it be:-
(5x - 2)^(1/3) = 2
(5x - 2) = 8
5x = 10
x = 2
2007-05-08 08:50:41 · answer #11 · answered by Como 7 · 0⤊ 0⤋