Vtyjkyuk

I recently stumbled upon the problem $3sqrtx-1+sqrt3x+1=2$, where I am supposed to solve the equation for x. My problem with this equation though, is that I do not know where to start in order to be able to solve it. Could you please give me a hint (or two) on what I should try first in order to solve this equation?

Note that I only want hints.

Thanks for the help!

HINT

We have

$$sqrt a + sqrt b=c stackrelboth , terms, ge 0iff (sqrt a + sqrt b)^2=a+2sqrtab+b=c^2 $$

and

$$a+2sqrtab+b=c^2 colorredimplies (2sqrtab)^2=(c^2-a-b)^2$$

for the latter implication we need to check at the end for possible extra solutions.

Similar to what others have said but I think a little simpler: write $3sqrtx- 1+ sqrt3x+ 1= 2$ as $3sqrtx- 1= 2- sqrt3x+ 1$ and square both sides: $9(x- 1)= 4- 4sqrt3x+ 1+ 3x+ 1$. Now write that as $6x- 14= -4sqrt3x+ 1$ and square again: $36x^2- 168x+ 196= 16(3x+ 1)= 48x+ 16$.

$36x^2- 120x+ 180= 0$.

$3x^2- 10x+ 15= 0$.

Of course, squaring twice might have introduced "spurious" solutions so solve that quadratic equation and check the solutions in the original equation.

Hint: Try the substitution $u=sqrtx-1$.

Added later: The answer hints so far, including mine (above), have all aimed at squaring away the square root symbols, leaving a quadratic equation that's easy to solve, with the caveat that the solutions to the quadratic are not necessarily solutions to the original equation. That approach works for any equation of the form $Asqrtax+b+Bsqrtcx+d=C$. But for this specific equation, it turns out there's an easy solution. Since the OP has asked only for hints, I'll state the key idea in the form of a question:

What can you say about $3sqrtx-1+sqrt3x+1$ if $xgt1$?

First step

$$3sqrtx-1+sqrt3x+1=2implies (3sqrtx-1+sqrt3x+1)^2=2^2.$$

Second step

After rearranging you'll get

$$6sqrtx-1sqrt3x+1=ax+b.$$ Take squares one more time.

Final step

Solve the quadratic equation and check that the solutions you get are solutions of the initial equation.

Let $A:=sqrt3x+1$ and $B:=sqrt3x-3$. Then, $A+sqrt3B=2$ and $A^2-B^2=4$. That is,
$$(A+sqrt3B)^2=A^2-B^2,.$$

Suppose that we are solving over the reals. Thus, $left(2sqrt3A+4Bright),B=0$. Since $A$ and $B$ are nonnegative and cannot simultaneously be $0$, we conclude that $B=0$.

Alternatively, we note that $xgeq 1$ so that $sqrtx-1$ is a real number. Thus, $$sqrt3x+1geq sqrt3cdot 1+1=2,.$$ For the required equality to hold, we must then have $sqrt3x+1=2$. Therefore, ....

Put $f(x) := 3sqrtx-1+sqrt3x+1$ on $[1,infty)$. Then $f'(x) > 0$ and $f(1) = 2$.

Hint :

Let,$$x=t^2+1$$
So, $$3t+sqrt3t^2+4=2$$
$$implies 3t^2+4=(2-3t)^2$$

Now,it's your turn to go on...

The answer should be $$x=1,if~t=0$$

$$and$$ $$x=5,if~t=2$$