For equations in which the variable(s) is/are under a radical.
Questions tagged [radical-equations]
89 questions
3
votes
5 answers
Solving $(x+1)\sqrt{2(x^2 + 1)} + \sqrt{6x^2 + 18x +12} = \frac{3x^2 + 7x + 10}{2}$
Today, I came across this problem.
$$(x+1)\sqrt{2(x^2 + 1)} + \sqrt{6x^2 + 18x +12} = \dfrac{3x^2 + 7x + 10}{2}$$
We are asked to find the possible values of $x$ satisfying this equation.
The first thought which came to my mind is to use some kind…
user983440
2
votes
8 answers
Solve the equation $x+\frac{x}{\sqrt{x^2-1}}=\frac{35}{12}$
Solve the equation $$x+\dfrac{x}{\sqrt{x^2-1}}=\dfrac{35}{12}.$$
The equation is defined for $x\in\left(-\infty;-1\right)\cup\left(1;+\infty\right).$ Now I am thinking how to get rid of the radical in the denominator, but I can't come up with…
kormoran
- 2,963
2
votes
1 answer
Need help solving the following equation
I'm having difficulty solving the following equation for $x$:
$$\sqrt[m]{(1+x)^2}-\sqrt[m]{(1-x)^2}=\sqrt[m]{1-x^2}$$
I have tried a few substitutions but that didn't seem to get me anywhere. Can someone please help me with that? Thanks in advance.
daniels
- 485
- 2
- 11
1
vote
2 answers
Are there any simpler ways to determine the solution for $\sqrt{x+\sqrt{x}}=1$ without back substitution checks?
A weak condition by inspection: $x>0$.
\begin{gather}
\sqrt{x+\sqrt x} = 1\\
x+\sqrt x = 1\\
\sqrt x = 1-x\\
x = 1-2x+x^2\\
x^2 - 3x + 1 =0\\
x=\frac{3\pm\sqrt5}{2}
\end{gather}
As both satisfy the weak condition $x>0$ it seems to me both are the…
D G
- 351
- 1
- 11
1
vote
3 answers
A golden question $\sqrt{2+\sqrt{2-x}}=\sqrt{x-\frac 1x} + \sqrt{1-\frac 1x}$
How would you solve this problem for real $x$?
$$\sqrt{2+\sqrt{2-x}}=\sqrt{x-\frac 1x} + \sqrt{1-\frac 1x}$$
It can be easily shown that both equations
$$x=\sqrt{2+\sqrt{2-x}}\tag{1}$$
and $$x=\sqrt{x-\frac 1x} + \sqrt{1-\frac 1x}\tag{2}$$
have the…
Hypergeometricx
- 22,657
1
vote
1 answer
simplify the equation $\frac{36}{\sqrt{x}} + \frac{9}{\sqrt{y}} = 42-9\sqrt{x}-\sqrt{y}$
This is the question:
$$\frac{36}{\sqrt{x}} + \frac{9}{\sqrt{y}} = 42-9\sqrt{x}-\sqrt{y}$$
This is from a timed competition, and I would like to know the fastest way to do it. I'm not sure, but is there a faster way than to replace $\sqrt{x}$ and…
user807252
1
vote
2 answers
Rationalizing radical expressions using conjugates - How does this step work?
This is the full solution given in my book:
Can someone please explain to me how it goes from Step 4 to Step 5? Specifically, I do not understand how the numerator simplifies to -1 and how the first term in the denominator, (x-1) becomes x.
0
votes
1 answer
How to solve this radical equation for x?
Question:
$$\frac{x}{\sqrt{x^2+1}} = x^4 - x$$
I tried:
$$\rightarrow \frac{1}{\sqrt{x^2+1}} = x^ 3 - 1$$
$$\to\frac{\sqrt{x^2 + 1}+1}{\sqrt{x^2+1}} = x^3$$
Now rationalising it
$$\to \frac{x^2 +1-1}{x^2+1-\sqrt{x^2+1}} =…
user983440
0
votes
1 answer
Square root with rational exponent
It might seem very stupid question.
If $x^2=9$ then to solve for $x$ we take both principal $n$-th root of $9$, i.e. $3$ and the negative $n$-th root of $9$, i.e. $-3$. This is right until I found about rational exponents.
If I try to solve the…
ahmed allam
- 151
0
votes
1 answer
How to solve this equation with rational exponents?
I've been really struggling to solve this one, could you provide how you'd solve it?
$$3x^{2/3} + 4x^{1/3} =4$$
P97
- 43
0
votes
3 answers
Equations involving squaring a variable under a radical sign
If anyone can help me with how to go about solving these kind of equations i would really appreciate it. :-)
$$\sqrt{36-2x^2} = 4$$
Solve for X
-2
votes
2 answers
Difficult Radical Equation
I am stuck in the following radical equation
$$
(4x-1)\sqrt{x^3+1}=2x^3+2x+1.
$$
SOLUTION I have tried based on the guidance of the commenters and I think that the following solution is rather…
Blind
- 1,116