a math problem(from ninth grade)?

Question

2x/(x• x+1)-(x• x+1)/x=1

Answer 1

I assume you're supposed to solve for x? There are a few ways to approach this, but here's how I'd do it:

First you want to get rid of the fractions by multiplying by the denominators of both (one at a time).

Starting with:
2x/(x• x+1)-(x• x+1)/x=1

Multiply by (x• x+1):
2x - [(x• x+1) (x• x+1)]/x = (x• x+1)

Then multiply by x:
2x•x - [(x• x+1) (x• x+1)] = (x• x+1)•x

Now the fractions are gone. Next multiply out the terms.

(x• x+1) (x• x+1) = (x^2 + 1)^2 (^2 just means squared)
So it's just squaring a binomial expression. Remember that (a+b)^2 = a^2 + 2ab + b^2
Just substitute x^2 for a and 1 for b, to get
(x^2)^2 + 2(x^2)(1) + 1^2 = x^4 + 2x^2 + 1
(x^4 means x to the 4th power)

For (x• x+1)•x, you just have to multiply both terms by x, so you get: x^3 + x

So putting those back into the whole equation, you'll have:
2x - (x^4 + 2x^2 + 1) = x^3 + x

Get rid of the parentheses in on the right side by distributing the negative sign:

2x^2 - x^4 - 2x^2 - 1 = x^3 + x

Then collect all the terms on one side:

2x^2 - x^4 - 2x^2 - 1 - x^3 - x = 0

Collect terms with equal powers of x, which in this case is just 2x^ and -2x^2, and add them together:

-x^4 - 1 - x^3 - x = 0

I find it usually helps to put the terms in order by the power of the variable, so you'd have:

-x^4 - x^3 - x -1 = 0

It would help to multiply everything by -1, to make coefficients positive::

x^4 + x^3 + x +1 = 0

Now you get to solve that. You can factor out x^3 from the first two terms, to get:

x^3(x+1) + (x+1) = 0

Now you can factor out the (x+1), to get:

(x+1) (x^3+1) = 0

So either x+1 = 0, which means that x = -1, or
x^3 + 1 = 0, so x = 1 or -1.

Since we multiplied by x a few times, it's possible that we added a few solutions in there, so now check those solutions in the original equation.

If x = -1,
2x/(x• x+1)-(x• x+1)/x = [2(-1)]/[(-1)(-1)+1] - [(-1)(-1)+1]/(-1)
=(-2)/2 - (2)/(-1)
= -1 + 2
=1
So x = -1 is a solution.

If x = 1,
2x/(x• x+1)-(x• x+1)/x = 2/(1+1) - (1+1)/1
= 2/2 - 1/1
= 1- 1
=0
So x = 1 is not a solution. It's an extra root that showed up from multiplying by x (or things that included x).

So the only solution of the original equation is x = -1.

I'm not surprised you had trouble with that - it's ridiculously complicated for a 9th grade math class. I'm in college and it took me like an hour to solve that.

Answer 2

2x/(x• x + 1) - (x• x + 1)/x = 1
2x/(x^2x + 1) - (x^2 + 1)/x = 1
(2x^2 - (x^2 + 1)^2)/(x(x^2 + 1) = 1
(2x^2 - x^4 - 2x^2 - 1) / (x^3 + x) = 1
- x^4 - 1 = x^3 + x
x^4 + x^3 + x + 1 = 0
x^3(x + 1) + (x + 1) = 0
(x + 1)(x^3 + 1) = 0
(x + 1)(x + 1)(x^2 - x + 1) = 0
x = -1, (1 ± √(1 - 4))/2

Answer 3

a=(x*x+1)/x
so your ecuation is 2*a-1/a=1. |*a(multiply by a the whole ecuation) => 2*a*a-1=a=>2*a*a-a-1=0; so the roots are:
a=(1+ or - sqrt(1+4*2*1))/2
so one solution is a=(1+3)/2=2
and one solution is a=(1-3)/2=-1
we take the first root of the a ecuation:
a=2
a=(x*x+1)/x
so we have 2*x=x*x+1=> x*x-2*x+1=0=>we sould have 2 roots but i this case because it's a square (using formula a*a+b*b-2a*b=(a-b)*(a*b) they are equal root1=root2=
(x-1)^2=0 =>x-1=0=> x=1;

we take the 2nd root of the a ecuation:
a=-1
a=(x*x+1)/x
so we have -x=x*x+1=> x*x+x+1=0=> we have 2 roots;
x=( - 1+ or - sqrt(1-4))/2
we don't have real root numbers. if x is a complex number then:
root 1: x= (-1+i*sqrt(3))/2
root 2: x=(-1-i*sqrt(3))/2
where i in coplex numbers is sqrt(-1)

Answer 4

No. Euclid's third postulate is "geometry is for 10th grade and above". commonly i come across algebra is taught in ninth grade, and that geometry is for 10th grade except you're interior the stepped forward math music that objectives for calculus by way of senior twelve months.

Answer 5

but don't you think it's too complicated that way?
suppose a=(x^2+1)/x,then it's:2/a-a=1, 2-a^2=a, (a-1)(a+2)=0, so a=1or -2.
and when (X^2+1)/x=1, there's no answer, so only a=-2 works. the answer would be(x^2+1)/x=-2
x=-1

Answer 6

The algebra is messy to show in this format.
Multiple both sides by x^2+1
Then multiple both sides by x

Ans x= -1

Answer 7

2/2-2
2/0