Solving for x: Simple Algebraic Equation with Square Roots

[VictoriaV]

1. Homework Statement

Solve for x: (x-1)²(x+1)= √(2-2x²)

2. Homework Equations

3. The Attempt at a Solution
+/- ((x-1)²(x+1) (x-1)²(x+1)) = -2(x²-1)
+/- ((x-1)(x-1) (x+1)(x-1)²(x+1)) = -2
(x+1)(x-1)
+/- ((x-1)³(x+1)) = -2
+/- ((x²-2x+1)(x-1)(x+1)) = -2
+/- (x4 -2x³- 2x -1 ) = -2
x4 – 2x³ -2x +1 =0 or -x4 + 2x³ + 2x +3=0I feel like I'm using the wrong method and cannot reach an answer please help

 

[phinds]

You have to show your work so we can see where you are stuck, otherwise you are just asking us to solve it for you, which is not how this forum works.

 

[Mark44]

1. Homework Statement

Solve for x: (x-1)²(x+1)= √(2-2x²)

2. Homework Equations

3. The Attempt at a Solution
+/- ((x-1)²(x+1) (x-1)²(x+1)) = -2(x²-1)

What's with the +/- ? When you square both sides you don't get \pm something.

+/- ((x-1)(x-1) (x+1)(x-1)²(x+1)) = -2
(x+1)(x-1)

It looks like you divided both sides by (x + 1)(x - 1). That's not a good thing to do, as you are losing two of your solutions; namely, x = 1 and x = -1.

Instead of dividing both sides by (x + 1)(x - 1), add 2(x + 1)(x - 1) to both sides (getting 0 on the right side), and then factor the left side.

+/- ((x-1)³(x+1)) = -2
+/- ((x²-2x+1)(x-1)(x+1)) = -2
+/- (x4 -2x³- 2x -1 ) = -2
x4 – 2x³ -2x +1 =0 or -x4 + 2x³ + 2x +3=0


I feel like I'm using the wrong method and cannot reach an answer please help

 

[Ray Vickson]

1. Homework Statement

Solve for x: (x-1)²(x+1)= √(2-2x²)

2. Homework Equations

3. The Attempt at a Solution
+/- ((x-1)²(x+1) (x-1)²(x+1)) = -2(x²-1)
+/- ((x-1)(x-1) (x+1)(x-1)²(x+1)) = -2
(x+1)(x-1)
+/- ((x-1)³(x+1)) = -2
+/- ((x²-2x+1)(x-1)(x+1)) = -2
+/- (x4 -2x³- 2x -1 ) = -2
x4 – 2x³ -2x +1 =0 or -x4 + 2x³ + 2x +3=0

I feel like I'm using the wrong method and cannot reach an answer please help

Don't write +- in the first line. When you square the left-hand-side of your original equation, you always get a quantity ≥ 0. Also, by convention, √ always means the non-negative square root.

The right-hand-side only makes sense if $x^2 \leq 1$; otherwise you would be taking the square-root of a negative quantity, and that would not give a real number anymore.

Use the "rational root theorem" to look for simple values of x that solve your 4th order equation---but first, get the right equation to look at. That means going back to the start and doing correct operations at each step. You will end up with a single equation, not a pair of different alternative equations.

 

[tdreceiver17]

i found it really easy to just multiply it all out and put it in your calculator. you will see that it only has 2 x intercepts