# Solve the equation $\sqrt{3x-2} +2-x=0$

Solve the equation: $$\sqrt{3x-2} +2-x=0$$

I squared both equations $$(\sqrt{3x-2})^2 (+2-x)^2= 0$$

I got $$3x-2 + 4 -4x + x^2$$

I then combined like terms $x^2 -1x +2$

However, that can not be right since I get a negative radicand when I use the quadratic equation.

$x = 1/2 \pm \sqrt{((-1)/2)^2 -2}$

The answer is 6

• Remember that in general $\;(a+b)^2\neq a^2+b^2\;$ : your squaring isn't corret. Commented May 22, 2013 at 8:01

You can't square the equation the way that you did.

What you did was essentially $$\sqrt{3x-2}^2+(2-x)^2 = 0$$ but it should be $$\sqrt{3x-2}^2 = (x-2)^2$$ which becomes $$3x-2 = x^2-4x+4$$ Solve from there.

And generally, if you have an equation, you can't apply operations to only part of one side. The rule is that "what you do to one side, you do to the other". So if you square the left side, you square the right side. If you were to square the left side of your original equation, it would become $$(\sqrt{3x-2}+2-x)^2=0^2$$ which won't get you where you want to go (at least, not as easily - you can get there by using the original equation to substitute out the square root from that point).

• Wow! Thank you so much. You explained the process and steps thoroughly. Commented May 22, 2013 at 8:05
• I got two answers x1 = 3.5 + 2.5 = 6 and x2 = 3.5 - 2.5 = 1 Is there a reason why the answer book only has one answer rather than two? Commented May 22, 2013 at 8:15
• Because one of the two solutions doesn't correspond to a solution of the original equation. Have a look - when $x=1$, you have $\sqrt{1}+2-1 = 2\neq 0$. This happens because squaring removes a negative sign - $x=1$ is a solution to $\sqrt{3x-2}-2+x=0$. Commented May 22, 2013 at 8:32

What you did was not squaring both sides of the equation. If we have an equation $$a+b=0,$$ then squaring both sides produces \begin{align*} (a+b)^2&=0^2\\a^2+2ab+b^2&=0. \end{align*} In your case, you have $a=\sqrt{3x-2}$ and $b=2-x$. But when you squared the left side of the equation, what you've written is instead $a^2+b^2$. But $$(a+b)^2=a^2+2ab+b^2\neq a^2+b^2.$$ Thus, when correctly squaring both sides of the equation you've been given, the result is $$(\sqrt{3x-2})^2 +2(\sqrt{3x-2})(2-x)+(2-x)^2=0.$$ However, this is not a productive place to start solving the question. A better approach would be to move the things not inside the radical to the other side, and then square, like this: \begin{align*} \sqrt{3x-2} +2-x&=0\\\\ \sqrt{3x-2} &=x-2\\\\ \qquad\qquad\qquad(\sqrt{3x-2})^2&=(x-2)^2\qquad \leftarrow \text{squared both sides} \end{align*} Can you take it from here?

• Yes! Thank you so much. Commented May 22, 2013 at 8:10

$$\sqrt{3x-2} +2-x=0$$

Isolating the radical:$$\sqrt{3x-2} =-2+x$$

Squaring both sides:$$\bigg(\sqrt{3x-2}\bigg)^2 =\bigg(-2+x\bigg)^2$$

Expanding $(-2+x)^2$ and gathering like terms: $$3x-2=-2(-2+x)+x(-2+x)$$

$$3x-2=4-2x-2x+x^2$$

Set x equal to zero:$$3x-2=4-4x+x^2$$

Gather like terms:$$0=4+2-3x-4x+x^2$$

Factor the quadratic and find the solutions:$$0=x^2-7x+6$$

$$0=(x-6)(x-1)$$

$$0=x-6\implies\boxed{6=x}$$ $$0=x-1\implies\boxed{1=x}$$

Checking 6 as a solution:

$$\sqrt{3(6)-2} +2-(6)=0$$

$$\sqrt{16} +2-6=0$$

$$4+2-6=0$$

$$6-6=0$$

Checking 1 as a solution:

$$\sqrt{3(1)-2} +2-(1)=0$$

$$\sqrt{1} +2-1=0$$

$$2\neq0$$

The solution $x=1$ does not equal zero and therefore is not a solution.

The solution $x=6$ does equal zero and therefore is our only solution to this equation.

If you meant

$$\sqrt{3x-2}+2-x=0\implies \sqrt{3x-2}=x-2\implies3x-2=(x-2)^2=x^2-4x+4\implies$$

$$\implies x^2-7x+6=0$$

Now just check that $x^2-7x+6=(x-6)(x-1)$ ...and remember to check at the end whether both solutions of this quadratic are actually solutions of your original equation, since when squaring some mess can happen there. For example, one of the soltuions of the quadratic is not a solution of your equation.