i need help in solving this question: x^2 +x^2/(x+1)^2=1?

Answer 1

Multiply by (x+1)²:
x²(x+1)² + x² = (x+1)²
Expand:
x²(x²+2x+1) + x² = x²+2x+1
x^4 + 2x³ + 2x² = x² + 2x + 1
Subtract x²+2x+1:
x^4 + 2x³ + x² - 2x - 1 = 0
Test possible rational roots (here using synthetic division):

1| 1 2 1 -2 1
...... 1 3 ..4 2
-----------------
... 1 3 4 ..2 3

So 1 is not a root.

-1| 1 2 1 -2 1
...... -1 -1 0 2
------------------
.....1 1 0 -2 3

So -1 is not a root. However, 1 and -1 are the only possible rational roots of this equation, so this equation has no rational roots. Therefore we must resort to the general method for solving quartics. Recall the equation:

x^4 + 2x³ + x² - 2x - 1 = 0

We convert this into a depressed quartic by making the substitution x=(t-1/2):

(t-1/2)^4 + 2(t-1/2)³ + (t-1/2)² - 2(t-1/2) - 1 = 0

t^4 - 2t³ + 3t²/2 - t/2 + 1/16 + 2(t³ - 3t²/2 + 3t/4 - 1/8) + (t² - t + 1/4) - 2(t-1/2) - 1 = 0

t^4 - t²/2 - 2t + 1/16 = 0

Now, the job is done if we can factor this into two quadratics. We know that for some constants a, b, c, and d, we have:

t^4 - t²/2 - 2t + 1/16 = (t² + at + b)(t² + ct + d)

Expanding the right hand side, we have:

t^4 - t²/2 - 2t + 1/16 = t^4 + (a+c)t³ + (b+ac+d)t² + (ad + bc)t + bd

Thus:

a+c=0
b+ac+d = -1/2
ad+bc=-2
bd=1/16

Since a+c=0, a=-c, so:

b-c²+d = -1/2
(b-d)c = -2
bd = 1/16

Simplifying the first two equations:

b+d = -1/2+c²
b-d=-2/c

Squaring both sides:

b²+2bd+d² = c^4 - c² + 1/4
b²-2bd+d² = 4/c²

Subtracting these equations:

4bd = c^4 - c² + 1/4 - 4/c²

And since bd = 1/16, we have:

1/4 = c^4 - c² + 1/4 - 4/c²

Since the only terms here are c², we make the substitution z=c². Then this becomes:

1/4 = z² - z + 1/4 - 4/z

Subtracting 1/4:

0=z² - z - 4/z

Multiplying by z:

z³ - z² - 4 = 0

We have therefore managed to reduce this quartic equation to a cubic equation. There's a general method for solving cubics, but it's at least as complicated as what we've done so far, so let's look for some rational roots first. In particular, let's try z=2:

2| 1 -1 0 -4
....... 2 2 ..4
---------------
... 1 1 2 ...0

So z³ - z² - 4 = (z²+z+2)(z-2) and 2 is a root. The other two roots would be easy to find at this point, but to get the factorization of the original equation, it was only necessary to find one root. With z=2, we go back to our system of equations:

a+c=0
b+ac+d = -1/2
ad+bc=-2
bd=1/16

Since we set c²=z, and z=2, we have c=√2, which gives a=-√2. Now recall that during the course of solving this system, we obtained the simplification:

b+d = -1/2+c²
b-d=-2/c

Adding gives:

2b = -1/2+c²-2/c
b=-1/4 + c²/2 - 1/c
Substituting c=√2:
b=3/4 - 1/√2

Similarly, subtracting the two equations gives:

2d = -1/2+c²+2/c
d = 3/4 + 1/√2

So we have:

t^4 - t²/2 - 2t + 1/16 = (t² - √2t + 3/4 - 1/√2)(t² + √2t + 3/4 + 1/√2)

You may confirm that this factorization is valid by manually multiplying out the right-hand side.

Now we merely need to solve two quadratic equations to get the roots of t^4 - t²/2 - 2t + 1/16 -- we will do this using the quadratic formula:

roots of (t² - √2t + 3/4 - 1/√2):

t=(√2 ± √(2-(3-4/√2))/2 = (√2 ± √(4/√2-1))/2

roots of t² + √2t + 3/4 + 1/√2:

t=(-√2 ± √(2-(3+4/√2))/2 = (-√2 ± i√(4/√2+1))/2

Finally, remember the substitution we made in the original polynomial -- x=(t-1/2). So we must subtract 1/2 from each of these answers. The roots of x^4 + 2x³ + x² - 2x - 1 = 0, and thus the possible solutions of x^2 +x^2/(x+1)^2=1, are then:

x = (√2 - 1 ± √(4/√2-1))/2 or x = (-√2 - 1 ± i√(4/√2+1))/2

Edit: corrected typo in final answer

Answer 2

question selection a million : For this equation x^2 - x - 12 = 0 , answer right here questions : A. locate the roots utilising Quadratic formulation ! answer selection a million : The equation x^2 - x - 12 = 0 is already in a*x^2+b*x+c=0 type. via matching the consistent place, we are able to derive that the value of a = a million, b = -a million, c = -12. 1A. locate the roots utilising Quadratic formulation ! remember the formulation, x1 = (-b+sqrt(b^2-4*a*c))/(2*a) and x2 = (-b-sqrt(b^2-4*a*c))/(2*a) We had understand that a = a million, b = -a million and c = -12, we basically ought to subtitute the value of a,b and c interior the abc formulation. So x1 = (-(-a million) + sqrt( (-a million)^2 - 4 * (a million)*(-12)))/(2*a million) and x2 = (-(-a million) - sqrt( (-a million)^2 - 4 * (a million)*(-12)))/(2*a million) which may be became into x1 = ( a million + sqrt( a million+forty 8))/(2) and x2 = ( a million - sqrt( a million+forty 8))/(2) Which make x1 = ( a million + sqrt( forty 9))/(2) and x2 = ( a million - sqrt( forty 9))/(2) We have been given x1 = ( a million + 7 )/(2) and x2 = ( a million - 7 )/(2) So we've the solutions x1 = 4 and x2 = -3

Answer 3

x^2 +x^2/(x+1)^2=1
x^2 [ (x+1)^2 + 1] = (x+1)^2
x^2 [x^2 + 2x + 1 + 1 ] = x^2 + 2x + 1
x^2 [x^2 + 2x + 2 ] = x^2 + 2x + 1
x^4 + 2x^3 + 2x^2 = x^2 + 2x + 1
x^4 + 2x^3 + 2x^2 - x^2 - 2x - 1 = 0
x^4 + 2x^3 + x^2 - 2x - 1 = 0

Answer 4

x^2 +x^2/(x+1)^2=1
x^2 ( (x+1)^2 + 1) = (x+1)^2
x^2 (x^2 + 2x + 1 + 1 ) = x^2 + 2x + 1
x^2 (x^2 + 2x + 2 ) = x^2 + 2x + 1
x^4 + 2x^3 + 2x^2 = x^2 + 2x + 1
x^4 + 2x^3 + 2x^2 - x^2 - 2x - 1 = 0
x^4 + 2x^3 + x^2 - 2x - 1 = 0

Answer 5

x² + x²/(x+1)² = 1
x²(x+1)² + x² = (x+1)²
x²(x² + 2x +1) + x² = x² + 2x +1
x^4 + 2x³ + x² + x² = x² + 2x +1
x^4 + 2x³ + x² - 2x - 1 = 0

Answer 6

X^2+X^2/(X+1)^2=1
X^2/(X+1)^2=1-X^2
X^2=(1-X^2)(X+1)^2=(X+1)^2-X^2(X+1)^2=X^2+2X+1-X^4-2X^3-X^2
X^2=-X^4-2X^3+2X+1
0=X^4+2X^3+X^2-2X+1...You can continue to solve this equation!!

Answer 7

Wow you've got all the help you could need here already!!

Answer 8

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the above is an image of the equation