# Simplify $\sqrt{3-2x^2+x}$: do I start by factoring or something else?

I have an expression: $$\sqrt{3-2x^2+x}$$ that I am to simplify. The provided solution is $$\sqrt{2}$$ and I do not see how to arrive at that.

I'm not sure how to start. Casting my mind back to previous chapters in my book I recall that with a radical expression I can re write using the product or quotient rules but I don't see ow I could use those here.

I tried writing the radicand as a factor: $$\sqrt{(x+1)(-2x+3)}$$ but after staring at this for a while I do not see what my next step is or if I should have even bothered to factor?

How can I simplify $$\sqrt{3-2x^2+x}$$ to arrive at $$\sqrt{2}$$?

I'm especially interested in the more granular, in between steps to get there (especially where to start).

[EDIT] Here is a screen shot of the question in my online textbook, in case I missed anything since from the comments it looks like the provided solution is incorrect. Note it's the second question that comes with an answer, please ignore the first one in the screen.:

• to arrive at $\sqrt2$, set $x=1$ ;-) Dec 10 '19 at 17:19
• Also $-0.5$ : )
– Ak.
Dec 10 '19 at 17:19
• The provided solution is $\sqrt{2}$ and I do not see how to arrive at that. --- We don't either, and in fact this is clearly incorrect. I suspect you have either neglected part of the question, incorrectly transcribed the question, there is an error/omission in the original statement, etc. Maybe you could tell us where this is from (e.g. which page of which book), as this might help. Dec 10 '19 at 17:25
• What? $\sqrt{3-2x^2 +x}\ne \sqrt 2$ unless $x=1$ or $x = -\frac 12$. If $x =$...... oh, say ,...$0$ then $\sqrt {3 - 2x^2 +x}=\sqrt 3$. Are you sure you weren't asked to evaluate for a specific value of $x$? Or to solve for $x$ if we are told that $\sqrt{3-x^2+x} = \sqrt 2$? Dec 10 '19 at 17:33
• Hi, I've added a screen shot of the question now, sorry if I missed something while posting. Dec 10 '19 at 17:55

Since $$\sqrt{3-2\times0^2+0}=\sqrt3\neq\sqrt2$$, your expression is not always equal to $$\sqrt2$$ (although sometimes it is).

• it is $\sqrt2$ when $x=1$ or $-\frac12$ Dec 10 '19 at 17:38

You can write $$-2\left(x^2-\frac{1}{2}x-\frac{3}{2}\right)$$ and this is $$-2\left(x^2-2\times \frac{1}{4}x+\frac{1}{16}-\frac{3}{2}-\frac{1}{16}\right)$$ this is $$-2\left(x-\frac{1}{4}\right)^2-2\left(-\frac{3}{2}-\frac{1}{16}\right)$$ this is equal to $$-2\left(x-\frac{1}{4}\right)^2+\frac{13}{4}$$ Now Substitute $$t=x-\frac{1}{4}$$

• What question is this an answer to? Dec 10 '19 at 17:50

I suspect the prime was that you were told to solve for $$x$$ IF we know that $$\sqrt{3-2x^2+x} = \sqrt 2$$.

$$\sqrt{3-2x^2+x}$$ is not a constant and takes many values different $$x$$ (although it is undefined for most ranges)and is only equal to $$\sqrt 2$$ at two of them.

$$\sqrt{3-2x^2 +x}=\sqrt 2$$ would only be true if $$3-2x^2 + x = 2$$ and it'd be a reasonable question to ask when that would occur.

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Although $$\sqrt{3-2x^2+x}=\sqrt{(x+1)(-2x+3)}$$ I wouldn't call that a "simplification".... although we can use that

a) solve for $$\sqrt{(x+1)(-2x+3)} =0$$. That occurs if $$x = -1$$ or $$x=\frac 32$$. (To equal $$0$$ then $$x+1 = 0$$ or $$-2x+3 =0$$

b) find out where $$\sqrt{(x+1)(-2x+3)}$$ is defined. It's only defined when $$-1 \le x \le \frac 32$$. (To be defined $$(x+1)(-2x+3) \ge 0$$ that happens when [$$x+1 \ge 0$$ and $$-2x+3 \ge 0$$] or if [$$x+1 \le 0$$ and $$-2x+3 \ge 0]$$... the second never occurs)

In the screen shot you have $$g\circ f (1)$$, not $$g\circ f(x)$$. So the last line plugs in $$1$$ for $$x$$.