# Questions tagged [radical-equations]

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28 questions
1answer
52 views

2answers
37 views

### Confusing regarding a nested radical equation

For all $a\in\Bbb R$ solve the equation $$\sqrt{x^2+4a^2\sqrt{x+a}}=x+2a$$ It is immediate to see that we got the restriction $x\geqslant-a$ (even though not given I assume that this equation is ...
1answer
55 views

### If $f(x) \Bbb Q[x]$ has splitting field of degree $16$ over $\Bbb Q$ , are the roots of $f(x)$ solvable by radicals.

If $f(x) \Bbb Q[x]$ has splitting field of degree $16$ over $\Bbb Q$ , are the roots of $f(x)$ solvable by radicals. My attemt: Any general equation of degree $\geq 5$ is not solvable by radicals. ...
3answers
173 views

2answers
48 views

### How to solve a limit through rationalizing but the radicand is a linear relation and the denominator is a quadratic? [closed]

The question is : $$\lim_{x\to 0}{\frac{\sqrt {3x+4}-6}{x^2}}$$ My friend and I are absolutely stumped and can only attempt to solve through substitution. Any ideas?
3answers
44 views

### Solving a radical function [closed]

I already did the previous algebra and i'm on this one step that I can't seem to get past. $$6^{\frac16}\cdot\frac{2x^8}{15}=2x^b$$ solve for b thank-you
4answers
91 views

### If $a+\sqrt{a^2+1}= b+\sqrt{b^2+1}$, then $a=b$ or not?

It might be a silly question but if $$a+\sqrt{a^2+1}= b+\sqrt{b^2+1},$$ then can I conclude that $a=b$? I thought about squaring both sides but I think it is wrong! Because radicals will not be ...
3answers
28 views

### Equations involving squaring a variable under a radical sign

If anyone can help me with how to go about solving these kind of equations i would really appreciate it. :-) $$\sqrt{36-2x^2} = 4$$ Solve for X
3answers
52 views

### Irrational equation

Solve over the real numbers: $$(x^2+x+1)^{1/3}+(2x+1)^{1/2}=2$$ I know for the second radical to be defined $x≥-0,5$ and I've attempted various methods I've solved other such equations with but to no ...
3answers
199 views

### Solve $\sqrt{\frac{4-x}x}+\sqrt{\frac{x-4}{x+1}}=2-\sqrt{x^2-12}$

The equation is $$\sqrt{\frac{4-x}x}+\sqrt{\frac{x-4}{x+1}}=2-\sqrt{x^2-12}$$ I tried squaring both left side and right side then bringing them to same numerator but got lost from there ... any ...
2answers
107 views