Very difficult equation

[MM]

Solve in real numbers the equation
[tex]|x+\sqrt{1-x^{2}}|=\sqrt{2}\left(2x^{2}-1\right)[/tex].

[srun]

See attachment.[/img]

[martosss]

Substitute [tex]x=cos\alpha, \alpha \in[0: ;\: \pi ][/tex]
[tex]Then\; |x+\sqrt{1-x^2}|=\sqrt{2}(2x^2-1) |cos\alpha +sin\alpha |=\sqrt{2}(2cos^2\alpha -1)[/tex]
[tex]|\N {\sqrt{2}}cos(\alpha -\frac{\pi}{4})|=\N {\sqrt{2}}cos(2\alpha )\ \alpha \in[0\: ;\: \frac{\pi}{4}]\cup [\frac{3\pi}{4}\: ;\: \pi][/tex]
1) [tex]\alpha \in [0\: ;\: \frac{\pi}{4}][/tex]
[tex]cos(\alpha -\frac{\pi}{4})=cos(2\alpha )\dots[/tex]
2. [tex]\alpha \in [\frac{3\pi}{4}\: ;\: \pi][/tex]
[tex]-cos(\alpha -\frac{\pi}{4})=cos(2\alpha )\dots[/tex]

[Guest]

Solve in real numbers the equation
[tex]|x+\sqrt{1-x^{2}}|=\sqrt{2}\left(2x^{2}-1\right)[/tex].

[Guest]

This equation can be written as "|x + 1 - x^2| = 2(2x^2 - 1)." To solve this equation, you can follow these steps:

Move all the terms to one side of the equation, so that the left-hand side is equal to zero. This will give you the equation "|x + 1 - x^2| - 2(2x^2 - 1) = 0."
Break the equation into two separate cases, one for when the expression inside the absolute value brackets is positive and one for when it is negative.
For the case where the expression inside the absolute value brackets is positive, you will have the equation "x + 1 - x^2 = 2(2x^2 - 1)."
For the case where the expression inside the absolute value brackets is negative, you will have the equation "-(x + 1 - x^2) = 2(2x^2 - 1)."
Solve each of these equations separately for x.
The solutions to the original equation will be the values of x that make both "x + 1 - x^2 = 2(2x^2 - 1)" and "-(x + 1 - x^2) = 2(2x^2 - 1)" true.

[Guest]

We can simplify the expression inside the absolute value as follows:

|x + (1 - x)/2| = 2/(2x^2 - 1)
Expanding the first term:

|x + 1/2 - x/2| = 2/(2x^2 - 1)
Combining like terms:

|1/2 - x/2| = 2/(2x^2 - 1)

The absolute value always returns a non-negative value, so we can drop the absolute value signs and consider the two possible cases:

1/2 - x/2 = 2/(2x^2 - 1)
or

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

Solving each equation separately and discarding extraneous solutions, we find:

x = -1
or

x = 1

So, the solution to the equation is x = -1 or x = 1.

[Guest]

-1/[tex]\sqrt{2}[/tex]

[Guest]

[tex]x^{2 }[/tex]

[Guest]

To solve the expression -1/sqrt(2), we need to simplify it further:

-1/sqrt(2) can be rewritten as -1 * (1/sqrt(2)).

Now, let's rationalize the denominator by multiplying both the numerator and denominator by sqrt(2):

-1 * (1/sqrt(2)) * (sqrt(2)/sqrt(2)) = -sqrt(2)/2.

Therefore, -1/sqrt(2) simplifies to -sqrt(2)/2.

[Guest]

Solve in real numbers the equation
[tex]|x+\sqrt{1-x^{2}}|=\sqrt{2}\left(2x^{2}-1\right)[/tex].

[Guest]

Substitute [tex]x=cos\alpha, \alpha \in[0: ;\: \pi ][/tex]
[tex]Then\; |x+\sqrt{1-x^2}|=\sqrt{2}(2x^2-1) |cos\alpha +sin\alpha |=\sqrt{2}(2cos^2\alpha -1)[/tex]
[tex]|\N {\sqrt{2}}cos(\alpha -\frac{\pi}{4})|=\N {\sqrt{2}}cos(2\alpha )\ \alpha \in[0\: ;\: \frac{\pi}{4}]\cup [\frac{3\pi}{4}\: ;\: \pi][/tex]
1) [tex]\alpha \in [0\: ;\: \frac{\pi}{4}][/tex]
[tex]cos(\alpha -\frac{\pi}{4})=cos(2\alpha )\dots[/tex]
2. [tex]\alpha \in [\frac{3\pi}{4}\: ;\: \pi][/tex]
[tex]-cos(\alpha -\frac{\pi}{4})=cos(2\alpha )\dots[/tex]

[RobertMills]

To solve the equation "|x + 1 - x^2| = 2(2x^2 - 1)," you can proceed as follows:

Consolidate the terms to one side of the equation to set the left-hand side to zero, yielding "|x + 1 - x^2| - 2(2x^2 - 1) = 0."

Divide the equation into two cases: one where the expression inside the absolute value brackets is positive and one where it is negative.

For the case where the expression inside the absolute value brackets is positive, you'll get "x + 1 - x^2 = 2(2x^2 - 1)."

For the case where the expression inside the absolute value brackets is negative, you'll have "-(x + 1 - x^2) = 2(2x^2 - 1)."

Solve each of these equations separately for x.

The solutions to the original equation will be the values of x that satisfy both "x + 1 - x^2 = 2(2x^2 - 1)" and "-(x + 1 - x^2) = 2(2x^2 - 1)."

Grasping the fundamentals is not so easy. It require a lot of practice with samples and assignments. I would suggest you to visit Maths assignment Help websites for their samples and assignment solution. You can also contact them at +1 (315) 557-6473.

[yandon]

i am a math teacher in china ,we can talk about math we can talk on wechat