Solving an equation with a cubic root and a square root. This is too radical!!!

This video is about solving an equation that has a cubic root and a square root in it.

[armacham]

Another way to do it: if you say A = xx - 3, then you can rewrite the equation as:

cubert(1 - a) + sqrt(a) = 1

And that's also a lot easier to work with than having x-squareds. for example you can say cubert(1 - a) = 1 - sqrt(a), cube both sides, and it's pretty straightforward to solve from there.

This trick really only works if the contents of the radicals have significant similarities. I think the trick used in the video (of having two variables, A and B) should work even when it's not easy to express both radicals using just 1 variable.

[ashishpradhan]

I don't know what to write but its an excellent approach. Keep this momentum man and make us fimiliar with this type of amezing question.

[keinKlarname]

I have written it now several times: from any of your vids I can learn some "trick" - thanks a lot for this!

[michaelpurtell]

i did what you did but avoided the B part by just substituting X^2 in terms of A in the second radical ; saved a couple of steps . This now my favorite channel .

[mputuchimezie]

You are doing a great job. Blackpenred pen talked about this channel and I am enjoying it

[elkincampos]

More easy.
Take u=√(x^2-3) then u>=0. u^2=x^2-3. -u^2=3-x^2. Therefore
1-u^2=4-x^2.
Thus
\cuberoot(1-u^2)+u=1.

\cuberoot(1-u^2)=1-u. Note that u=0 and u=1 satisfy the equation. Now suppose that u\not=0 and u\not=1.
Then cuberoot ((1+u)/(1-u)^2)=1.
Thus ... u=3.
We conclude u=0, 1 or 3 there is x=+-√(3) or x=+-2 or x=+-/2*√3.

[adandap]

Nice. The straight ahead algebraic solution isn't too bad here. You get a sextic, but there are no odd terms, so it's really a cubic in x^2 and isn't too hard to solve.

[JohnRandomness]

Really, using u = x² does simplify the writing and the checking. It makes the first method, that you rejected, doable without too much trouble.

[tapankumarchakraborty]

Thanks, explained very nicely.I request you to kindly red coloured explanation should be shown in white background for clear visibility.thanks again.

[manojsurya]

It's a nice approach, substitution are great for reducing such equations 👍

[pkmath]

Another great one! Keep up the good one!

[xyz]

do a single substitution of y = x^2 - 3 will be simpler and faster.

[jarikosonen]

Looks like very nice method.
Can you solve what is Re{(a+b*i)^(1/3)}? (Real part of cube root of complex number a+b*i)

[dzpismu]

After following you some time, I found this problem is so easy to solve. Thank you!

[Drk]

I thought of using u=(x^2 - 3)^(1/6), so I was expecting up to 6 Solutions.
Your approach was nice, i like it :)

[-basicmaths]

This isn't so complicated.
By putting x^2=. t^3+4 We get
Cube rut - t^3+√(t^3+1)=1
-t+√( t^3+1)=1
√( t^3+1)=1+ t
t^3+1=(1+ t)^3
t^3+1=t^3+3t^2+3t+1
3t^2+3t=0
3t(t+1)=0
t=0, t=-1
Now x^2=t^3+4
x^2=4
x=(2, -2)
x^2=t^3+4
x^2=3
x=(√3, -√3)

[JohnRandomness]

Let u = x². One obvious solution is u = 4, giving 0 + 1 = 1. The use of doing any work is finding the other solution(s), such as u = 12. This gives -2 + 3 = 1.

cubert(4-u) = 1 - sqrt(u-3)
4 - u = 1 - 3sqrt(u-3) + 3(u-3) - (u-3)*sqrt(u-3)
12 - 4u = (-1)4(u - 3) = (-1)sqrt(u-3)*(3 + u-3) = (-1)u*sqrt(u-3) --- Dividing by sqrt(u-3), so should check u = 3, which is another obvious solution.

4*sqrt(u-3) = u
16(u-3) = u²
0 = u² - 16u + 48 = (u - 4)(u-12)

The three solutions are therefore u = 4, 12, and 3. (x = ±sqrt(u) gives six solutions for x.)