# Radical equation solve $\sqrt{3x+7}-\sqrt{x+2}=1$. Cannot arrive at solution $x=-2$

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:

$$\sqrt{3x+7}-\sqrt{x+2}=1$$

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

$$(2x+4)^2=(2\sqrt{x+2})^2$$

$$4x^2+16x+16=4(x+2)$$ # 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:

$$x=\frac{-12\pm\sqrt{12^2-(4)(4)(14)}}{2(4)}$$

$$x=\frac{-12\pm{\sqrt{(144-224)}}}{8}$$

$$x=\frac{-12\pm{\sqrt{-80}}}{8}$$

$$x=\frac{-12\pm{i\sqrt{16}*i\sqrt{5}}}{8}$$

$$x=\frac{-12\pm{4i*i\sqrt{5}}}{8}$$

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

This is as far as I get:

$$\frac{-12}{8}\pm\frac{4\sqrt{5}}{8}$$

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

How can I arrive at -2?

• 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. – Thomas Andrews Jun 25 '19 at 14:31
• Tangent: always make sure to re-apply the initial restrictions to your final answer—here namely $3x+7\ge0$ and $x+2\ge0$ – gen-ℤ ready to perish Jun 25 '19 at 14:40
• @ChaseRyanTaylor, Avoid squaring whenever possible as it often introduces math.stackexchange.com/questions/1513842/extraneous-roots – lab bhattacharjee Jun 25 '19 at 14:46
• @labbhattacharjee Yes, I’m aware – gen-ℤ ready to perish Jun 25 '19 at 14:47
• The final quadratic should be $4x^2 + 12x + 8 = 0$. – NickD Jun 25 '19 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 $$(b+1)^2-3b^2=1$$

• Substitution makes it simpler indeed. – NoChance Jun 25 '19 at 14:44
• I agree. In my opinion it's a best solution. +1 – Michael Rozenberg Jun 25 '19 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 $$1$$.

• Except the later calculations look like the square was taken correctly. – Thomas Andrews Jun 25 '19 at 14:32

$$\sqrt{3x+7}-\sqrt{x+2}=1$$

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

$$y=x+2=0$$$$x=-2$$

$$y-1=0$$$$y=x+2=1$$$$x=-1$$