# linear equation? Calculus review?

I can't nail this problem...

The given equation is either linear or equivalent to a linear equation. Solve the equation. (If there is no solution, enter NO SOLUTION. If all real numbers are solutions, enter REALS.)

$$\sqrt{3x} + \sqrt{12}= \frac{x+3}{\sqrt{3}}$$

Thanks for helping if you can!

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Is that "sqrt3x" supposed to be $\sqrt3x$ or $\sqrt{3x}$? You can get the first by writing $\sqrt{3}x$ and the second by writing $\sqrt{3x}$. Instructions for writing mathematics on this site are here. –  MJD Aug 31 '14 at 18:49

1. Multiply each side of the equation by $\sqrt{3}$:

$3\sqrt{x}+6=x+3$

2. Subtract $x+3$ from each side of the equation:

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

3. Replace $\sqrt{x}$ with $t$:

$-t^2+3t+3=0$

4. Find the roots of the quadratic equation:

$t_{1,2}=\dfrac{-3\pm\sqrt{9+12}}{-2}\approx-0.79,3.79$

5. Replace $t$ with $\sqrt{x}$:

• $\sqrt{x}\approx-0.79 \implies x\not\in\mathbb{R}$
• $\sqrt{x}\approx3.79 \implies x\approx14.37$

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wow, so clean explanation. –  doniyor Aug 31 '14 at 19:49
@doniyor: Thank you :) –  barak manos Aug 31 '14 at 19:59

Here's the idea:

Multiply both sides by $\sqrt{3}$ to get $$3\sqrt{x}+6=x+3$$

Now let $x=y^2$ and solve for $y$ (and then for $x$).

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Since you say the equation is linear I'll assume it's $$\sqrt{3}x + \sqrt{12} = x + 3/\sqrt{3}$$ This can be simplified to $$\sqrt{3}x + 2\sqrt{3} = x + \sqrt{3}$$ from which it follows that $$(\sqrt{3}-1)x =-\sqrt{3} \\ x =-\frac{\sqrt{3}}{\sqrt{3}-1} \approx -2.366$$

EDIT. The question has now changed, but since the equation is stated to be linear I suspect it should be $$\sqrt{3}x + \sqrt{12} = \frac{x + 3}{\sqrt{3}}$$ to which the solution is $$x = -\tfrac{3}{2}$$

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Using $x=−2.366$ gives $6.128$ on the left side of the equation and $0.366$ on the right side of the equation, so this is clearly not the correct solution. –  barak manos Aug 31 '14 at 20:11
That's because the question changed after I posted the answer! –  MartinG Aug 31 '14 at 20:17
That's not right, I agree. –  barak manos Aug 31 '14 at 20:20
It's $\sqrt{3x}$ in the question, not $\sqrt{3}x$. –  barak manos Aug 31 '14 at 21:04