I am to solve $\sqrt{3x+7}-\sqrt{x+2}=1$ and the solution is provided as -2.

Since this is a radical equation with 2 radicals, I followed suggested textbook steps of isolating each radical and squaring:


$(3x+7=(1-\sqrt{x+2})^2$ # square both sides

(Use perfect square formula on right hand side $a^2-2ab+b^2$)

$3x+7=1^2-2(1)(-\sqrt{x+2})+x+2$ # lhs radical is removed, rhs use perfect square formula

$3x+7=1+2(\sqrt{x+2})+x+2$ # simplify

$3x+7=x+3+2\sqrt{x+2}$ # keep simplifying

$2x+4=2\sqrt{x+2}$ # simplify across both sides


$4x^2+16x+16=4(x+2)$ # now that radical on rhs is isolated, square both sides again

$4x^2+12x+14=0$ # a quadratic formula I can use to solve for x

For use int he quadratic function, my parameters are: a=4, b=12 and c=14:






$x=\frac{-12\pm{-4\sqrt{5}}}{8}$ #since $4i*i\sqrt{5}$ and i^2 is -1

This is as far as I get:


I must have gone of course somewhere further up since the solution is provided as x=-2.

How can I arrive at -2?

  • 1
    $\begingroup$ You've got the wrong value at the first step: $3x+7=(1-\sqrt{x+2})^2.$ The minus should be a plus: $3x+7=(1+\sqrt{x+2})^2.$ Later steps proceed as if you had the correct formula, though. $\endgroup$ – Thomas Andrews Jun 25 at 14:31
  • $\begingroup$ Tangent: always make sure to re-apply the initial restrictions to your final answer—here namely $3x+7\ge0$ and $x+2\ge0$ $\endgroup$ – gen-z ready to perish Jun 25 at 14:40
  • $\begingroup$ @ChaseRyanTaylor, Avoid squaring whenever possible as it often introduces math.stackexchange.com/questions/1513842/extraneous-roots $\endgroup$ – lab bhattacharjee Jun 25 at 14:46
  • $\begingroup$ @labbhattacharjee Yes, I’m aware $\endgroup$ – gen-z ready to perish Jun 25 at 14:47
  • $\begingroup$ The final quadratic should be $4x^2 + 12x + 8 = 0$. $\endgroup$ – NickD Jun 25 at 17:30

Hint: Writing your equation in the form $$\sqrt{3x+7}=1+\sqrt{x+2}$$ squaring gives $$3x+7=1+x+2+2\sqrt{x+2}$$ so $$x+2=\sqrt{x+2}$$ squaring again: $$(x+2)^2=x+2$$ Can you finish?


Let $\sqrt{3x+7}=a,\sqrt{x+2}=b$

$\implies a,b\ge0$ and $a-b=1$

and $a^2-3b^2=1$

Or $(b+1)^2-3b^2=1$

  • $\begingroup$ Substitution makes it simpler indeed. $\endgroup$ – NoChance Jun 25 at 14:44
  • $\begingroup$ I agree. In my opinion it's a best solution. +1 $\endgroup$ – Michael Rozenberg Jun 25 at 14:55

Your first step is wrong. It should be $$\sqrt{3x+7}-\sqrt{x+2}=1\implies\sqrt{3x+7}=1+\sqrt{x+2}$$ so we have $$3x+7=(1+\sqrt{x+2})^2$$ from which I think you can continue.

Note that as a check to your textbook solution, at $x=-2$, we get $$\sqrt{3(-2)+7}-\sqrt{-2+2}$$ which is indeed equal to $1$.

  • $\begingroup$ Except the later calculations look like the square was taken correctly. $\endgroup$ – Thomas Andrews Jun 25 at 14:32


$3x+7=(1 \color{red}{\mathbf{ \,-\, }}\sqrt{x+2})^2$ # square both sides

You want: $3x+7=(1 \color{blue}{\mathbf{ \, + \,}}\sqrt{x+2})^2$

Note: not only $x=-2$ solves this equation, also $x=-1$.


In addition to the possible typo $\sqrt{3x+7}=1+\sqrt{x+2}$ not $1-\sqrt{x+2}$ in the RHS, you made an arithmetic error later.

From $4x^2+16x+16=4(x+2)$, you should get $4x^2+12x+8=0$, not $4x^2+12x+14=0$.


There are actually two solutions: $x = -1; x = -2$ when you continue with the ``Can you finish?'' step of Dr. S.


The big error is that $4x^2+16x+16=4(x+2)$ is the same as $4x^2+12x+8=0.$ You somehow got $4x^2+12x+14=0.$ Did you treat $4(x+2)$ as the same as $4x+2?$ The equation $4x^2+12x+8=0$ has $x=-1$ and $x=-2$ as roots.

There's an earlier error where you write: $3x+7=(1-\sqrt{x+2})^2.$ The right side should be $(1+\sqrt{x+2})^2,$ but your later expansion somehow yields the correct value - so two errors led to a correct expression.

It's easier, when you have $2x+4=2\sqrt{x+2},$ if you divide by $2$ before squaring, and get: $x+2=\sqrt{x+2}.$

One quick way to simplify it from the start is to set $y=x+2.$ Then $3y+1=3x+7.$ So you have a slightly simpler equation:

$$\sqrt{3y+1}-\sqrt{y}=1\\ \sqrt{3y+1}=1+\sqrt{y}\\ 3y+1 = 1+2\sqrt{y}+y\\ 2y=2\sqrt{y}\\ y=\sqrt{y}\\ y^2=y\\ y=0,1$$

You have to go back and check each $y$ in the original equation, then take $x=y-2$ for each solution $y.$


Let x+2=y then:

$\sqrt {3y+1}=\sqrt y +1$

squaring both sides we get:

$3y+1=y+1=2\sqrt y$

$y=\sqrt y $$y^2-y=y(y-1)=0$

that gives:




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