really need help!

sqrt(3x^3-4x)=x+1
Square both sides to get
3x^3-4x = x^2+2x+1 or
3x^3-x^2-6x-1 = 0
Are you sure you have the problem correct? I see no "nice" answers.
What are you studying?
 
That's what I did.
3x^3-x^2-6x-1 = 0
What are you studying?????
Newton's method perhaps?
You posted in "Other Math" so I don't have a clue.
---------------
Gene
 
If it were 3x<SUP>2</SUP> there is a nice answer?
 
yeah it is 3x^2 my mistake can you help me solve the equasion i am competely lost! tahk you
 
Gene gave you a very clear way to do this type of equation.
Square both sides: 3x<SUP>2</SUP>−4x=x<SUP>2</SUP>+2x+1.
2x<SUP>2</SUP>−6x−1=0.
Can you solve that?
 
zachary, looks like you only want the answer...

can you solve this one for x:
x^2 - 4x + 3 = 0
:?:
 
yeah then it would be (x-1)(x-3)
i can so that much but this equasion does not work out and it is really bugging me i have an answer of x=3 but i feel that is wrong.
 
It's one you can't factorise. You'll need to apply the quadratic formula or complete the square to solve for x.

Because of the squaring of the square root at the beginning, extraneous (extra) solutions for x may have cropped up. You need to check your two values for x by plugging them into the left-hand side of the original equation and checking that it equals the right-hand side. (Extraneous solutions are possible, but not necessary.)
 
zachary1010 said:
this equasion does not work out and it is really bugging me i have an answer of x=3 but i feel that is wrong.
Click to expand...
2x^2−6x−1=0 is the equation; substitute your x=3:
18 - 18 - 1 = 0
-1 = 0 : why did you think x=3 was a solution?

USE THE QUADRATIC!
x = [6 +- sqrt(6^2 - 4(2)(-1))] / (2(2))

Since quadratic formula is x = [-b +- sqrt(b^2 - 4ac)] / (2a),
you can "rewrite" it this way: 2ax + b = +-sqrt(b^2 - 4ac).
It is then easier to solve. Just a method I like to use.
 
The answer is \(\displaystyle \frac{3}{2} \pm \frac{{\sqrt {11} }}{2}\).
 
You have \(\displaystyle \underbrace 2_ax^2 \;\underbrace { - 6}_bx\;\underbrace { - 1}_c = 0\) therefore,

\(\displaystyle \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}} = \frac{{6 \pm \sqrt {36 - 4(2)( - 1)} }}{4} = \frac{{6 \pm \sqrt {44} }}{4} = \frac{3}{2} \pm \frac{{\sqrt {11} }}{2}\)
 
x = [6 +- sqrt(6^2 - 4(2)(-1))] / (2(2))
=[6 +- sqrt(44)]/4 =
[6 +- 2sqrt(11)]/4 =
3/2 +- sqrt(11)/2
 
wow! i finally got it thank you so much i see how i need to divide i was simplifing the radiacal and then got confused thanks a lot and i am sure i will be back for more questuons! thanks!
 
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