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I have solved this equation: $$ \sqrt{3x}-\sqrt{4x-5}=0 $$

And I got that the solutions are: $$ x=\frac{5}{7}, x=5 $$

My question is, did I do it right, because $$ x=\frac{5}{7} $$ dont fit in the equation. I mean, if I put 7/5 in equation, I will not get the correct result.

This is how I solved the equation: $$ \sqrt{3x}-\sqrt{4x-5}=0\\ \sqrt{3x}=\sqrt{4x-5}\\ (\sqrt{3x})²=4x-5\\ |3x|-4x+5=0\\ for\ x<0\\ -3x-4x=-5\\ -7x=-5\\ 7x=5\\ x=\frac{5}{7}\\ for\ x>0\\ 3x-4x=-5\\ -x=-5\\ x=5\\ x=5\\ $$ Thank you very much!!!

Gerry Myerson
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depecheSoul
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    You found that "solution" when you assumed $x < 0$, hence it is not a solution. – njguliyev Oct 13 '13 at 11:36
  • I assumed that x<0 and x>0, because I had absolute value in the equation. – depecheSoul Oct 13 '13 at 11:38
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    Note that $(\sqrt{3x})^2$ is simply $3x$ with the additional condition that $x\ge 0$. And $(\sqrt{4x-5})^2$ is simply $4x-5$ with the additional condition that $4x-5\ge0$. Introducing $|3x|$ with the absolute value function here does not really make sense (and if so, you'd have to do that on both sides). - I guess you mixed this up with $\sqrt{x^2}=|x|$. – Hagen von Eitzen Oct 13 '13 at 11:39
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    Look at what happens when you use your method to solve a much simpler equation, $\sqrt x=1$. You would go $(\sqrt x)^2=1$, $|x|=1$, and you would get $x=\pm1$, when the correct answer is $x=1$. If you understand what's wrong in the simple problem, maybe you'll see the difficulty in the original problem. – Gerry Myerson Oct 13 '13 at 11:44
  • Thanks for your help everybody, but the main problem for me now is how to get rid of sqrt function in the equation, because any way I try, I get absolute value. – depecheSoul Oct 13 '13 at 11:50
  • The only reason you get an absolute value is because you insist on writing one where it has no business being. The square of the square root of $x$ is $x$, not $|x|$, and I don't understand why you think anything different. – Gerry Myerson Oct 13 '13 at 12:35

1 Answers1

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As long as we are working with real numbers, the expression $\sqrt a$ is by definition the unique nonnegative number $b$ with the property that $b^2=a$, provided it exists (that is, we need $a\ge 0$ in the first place).

Unfortunately you did not join your lines of derivation with either $\Rightarrow$ or $\Leftrightarrow$ to tell the reader if you think that what you did was an equivalence transformation or just a conclusion. However, at one step you went from $$\tag1 \sqrt{3x} =\sqrt{4x-5}$$ to $$\tag2 |3x| = 4x-5.$$ These lines, as they stand, can unfortunately only connected by "$\Rightarrow$", not by "$\Leftrightarrow$". According to what I wrote in the first paragraph, you could have written simply $$\tag{2a}3x=4x-5$$ instead of $(2)$ and it would still be a valid conclusion (but not an equivalence), saving you the trouble of working with absolute values and cases. If the sqaure roots of two numbers are equal, then these numbres themselves are also equal!

In fact we could turn this step of the derivation into an equivalence by adding the conditions that the oriignal radicands are nonnegative: $$\tag{2b}3x=4x-5\text{ and }3x\ge 0\text{ and }4x-5\ge0$$ or slicghtly simplified $$\tag{2c}3x=4x-5\text{ and }\frac54\le x.$$ Then we would indeed have an equivalence $(1)\Leftrightarrow(2c)$.

The introduction of absolute value would be justified if root and squaring occured in different order: $$ \sqrt{a^2}=|a|\qquad\text{for all }a\in\mathbb R$$ $$ (\sqrt a)^2=a\qquad\text{if }a\ge 0 \text{ (and the left side is undefined if }a<0\text{)}.$$

Of course, it is then still true that $(\sqrt a)^2=|a|$, but this makes things unnecessarily complicated if we have to consider only the case $a\ge0$ anyway.

This is what one could have done: $$\begin{eqnarray}\sqrt{3x}-\sqrt{4x-5}&=&0\\ \iff\sqrt{3x}&=&\sqrt{4x-5}\\ \implies 3x&=&4x-5\\ \iff 0&=&x-5\\ \iff 5&=&x\end{eqnarray} $$ So we find the only possible solution is $x=5$. Since the derivation involved a mere implication in one step, we need to check: $\sqrt{3\cdot 5}-\sqrt{4\cdot 5-5}=\sqrt{15}-\sqrt{15}=0$, OK.

Alternatively, one could avoid the checking phase by enforcing equivalence in all steps: $$\begin{eqnarray}\sqrt{3x}-\sqrt{4x-5}&=&0\\ \iff\sqrt{3x}&=&\sqrt{4x-5}\\ \iff 3x&=&4x-5\land 3x\ge 0\land 4x-5\ge0 \\ \iff 0&=&x-5\land \frac54\le x\\ \iff 5&=&x\land \frac54\le x\\ \iff 5&=&x.\end{eqnarray} $$ Then again, it is a good idea anyway to crosscheck all solutions one finds ;)

Hagen von Eitzen
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