Challenge Q: solve x + sqrt[1-x^2] = sqrt[2](2x^2-1) [calculator not allowed]

[RandomGuyThatNoobOnMath]

My math teacher wanted us to challange ourselves, and so she gave this question
( Cannot use calculator )



??????????
no idea , no idea , no idea

 

[BigBeachBanana]

My math teacher wanted us to challange ourselves, and so she gave this question
( Cannot use calculator )

View attachment 34723

??????????
no idea , no idea , no idea

Hint:
\(\displaystyle (x+\sqrt{1-x^2})(x-\sqrt{1-x^2}) = 2x^2-1\)

 

[Dr.Peterson]

My math teacher wanted us to challange ourselves, and so she gave this question
( Cannot use calculator )

View attachment 34723

??????????
no idea , no idea , no idea

What methods have you learned? Did you try the standard way to solve a radical equation? What happened?

I was able to solve it by hand (without hints), by solving a quartic equation using the rational roots theorem (after a simple substitution that rationalized the coefficients). Does that sound familiar?

 

[Four Muffins]

How do you rationalise a coefficient by substitution? I googled a bit, and couldn't find anything like sounded like that, and can't think of it myself. The only substitution method I know of is to replace a variable, like u = x^2. I thought of replacing the radicals with an x, but it seems like it would be hard to keep track of and I'm not sure it makes sense.

I tried this problem too, and got stuck in a bunch of ways. I tried multiplying by a conjugate, isolating radicals on one side in different ways, factoring the left side, squaring each side varying times, and I usually ended up with a polynomial with radical coefficients, sometimes with fractions as well.

The only method I found that might work was to square until there were no more radicals, and ended up at an 8th degree polynomial with three digit coefficients to use long division on, but brute force wasn't an interesting way to solve the problem, so I stopped.

(Normally I'd post my attempts, but there are ten pages and I don't imagine anyone wants to read all that. )

 

[Dr.Peterson]

How do you rationalise a coefficient by substitution? I googled a bit, and couldn't find anything like sounded like that, and can't think of it myself. The only substitution method I know of is to replace a variable, like u = x^2. I thought of replacing the radicals with an x, but it seems like it would be hard to keep track of and I'm not sure it makes sense.

You haven't shown what you got by squaring; it is at that point that you can see the substitution I did. (I made up the name for the process.)

You will see that alternate coefficients have [imath]\sqrt{2}[/imath], so replacing x with [imath]\frac{u}{\sqrt{2}}[/imath] will simplify that.

The only method I found that might work was to square until there were no more radicals, and ended up at an 8th degree polynomial with three digit coefficients to use long division on, but brute force wasn't an interesting way to solve the problem, so I stopped.

I'd like to see this part of the work, namely how you got 8th degree, rather than my 4th degree.

I tried this problem too, and got stuck in a bunch of ways. I tried multiplying by a conjugate, isolating radicals on one side in different ways, factoring the left side, squaring each side varying times, and I usually ended up with a polynomial with radical coefficients, sometimes with fractions as well.

Did you try BBB's suggestion? I didn't see that idea myself, but it is extremely helpful, and saves most of the work I did. Replace his RHS in your equation with his LHS.

 

[Four Muffins]

You haven't shown what you got by squaring; it is at that point that you can see the substitution I did. (I made up the name for the process.)

You will see that alternate coefficients have [imath]\sqrt{2}[/imath], so replacing x with [imath]\frac{u}{\sqrt{2}}[/imath] will simplify that.

Thank you, it might never have occurred to me to sub in a quotient variable. I did find an answer using your method, but it was rather different to using BBB's hint, so I'm not sure which to believe. Or if I understood the problem correctly.

Spoiler

I'd like to see this part of the work, namely how you got 8th degree, rather than my 4th degree.

I just realised I made a mistake in the third line, making squaring the root of 2 as 4, so the coefficients are wrong.

Spoiler

Did you try BBB's suggestion? I didn't see that idea myself, but it is extremely helpful, and saves most of the work I did. Replace his RHS in your equation with his LHS.

I did, though I'm not sure entirely what I was to do with it, I just sort of meandered about with it. A single value for x was not what I was expecting after all the previous work, though.

Spoiler

 

[Dr.Peterson]

When you got the 8th degree polynomial, it was by squaring before isolating the radical, so you had to square twice.

I don't have time to check everything, but in the last method, you made the mistake of dividing by an expression with x in it. How do you know that can't be zero? The best thing to do is to factor it out rather than dividing

 

[Four Muffins]

I also messed up with squaring the root of 2 as 4 again...

Thank you, I hadn't considered dividing with variables as troublesome.

 

[Dr.Peterson]

I also messed up with squaring the root of 2 as 4 again...

Thank you, I hadn't considered dividing with variables as troublesome.

In the last method, the quadratic is correct, but you made the usual arithmetic error after that. (Watch signs!)

Then, don't forget that when you square in the course of solving an equation, you have to check for extraneous roots. There will be a couple (by either method).