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Q)What is the value of $x$ when $\frac{1}{4\sqrt{x}}=\frac{-1}{5}$ ?

Ans) First of all let me tell that I know how to find the value of $x$ in the above equation.

$\frac{1}{4\sqrt{x}}=\frac{-1}{5}$ $\implies \sqrt{x}=-\frac{5}{4}$ $\implies x=(-\frac{5}{4})^{2}$ $\implies x=\frac{25}{16}$.

My doubt:

I can't understand that how is it possible that $\sqrt{\frac{25}{16}}=\frac{-5}{4}$ because we know that $\sqrt{x^{2}}=|x|$. Therefore, $\sqrt{\frac{25}{16}}=|\frac{5}{4}|=\frac{5}{4}$. Please help me out with this equation.

For e.g.- when we solve equations like:

Find the value of $x$ when $x^{2}=16$.

Generally for finding the value of $x$ we take square root on both sides. $x^{2}=16$ $\implies \sqrt{x^{2}}=\sqrt{16}$ $\implies |x|=4$

Therefore the values of $x$ are $+4$ and $-4$.

So, we generally don't face any problem to solve these types of equations.

user1551
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DEB SANKAR ROY
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    LHS is positive, whereas RHS is negative – Sine of the Time Jan 19 '24 at 07:39
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    You are correct that the is answer is that there is no solution. That happens sometimes when a test wants to assess your ability to think critically rather than blindly manipulate terms. – Ninad Munshi Jan 19 '24 at 07:41
  • Squaring is no equivalence transformation. Consider $x=3$ which obviously has only the solution $3$. If we square , we get $x^2=9$ which has an additional solution $-3$. Therefore , whenever we square , we have to check whether the original equation is actually satisfied. – Peter Jan 19 '24 at 07:43
  • While is true that $(\pm 1)^2=1$ we have that by convention $\sqrt 1=1$, therefore we have no solution on the reals in this case. – user Jan 19 '24 at 07:51
  • Of course no real solution. But they are complex solutions. – JJacquelin Jan 19 '24 at 08:21
  • This implies that $x$ doesn't have any real solution but there can be complex value of $x$. – DEB SANKAR ROY Jan 19 '24 at 09:35
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    " But they are complex solutions." Not really. If we take the $\sqrt{}$ symbol to mean any square root (all non-zero numbers have two square roots) then $\sqrt{x} = -\frac 54$ and $x = (-\frac 54)^2=\frac {25}{16}$ which is a real number. But $\sqrt{}$ doesn't mean any square root. It usually means the principal square root which by convention means the positive root (if one is pos and the other neg). Allowing complex numbers doesn't give us any more solutions. It's just that in comx anal we might have looser definition of what the $\sqrt{}$ symbol means. – fleablood Jan 19 '24 at 16:37
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    Note: $\sqrt{x^2} = |x|$ isn't really true. If $i$ is the imaginary unit so that $i^2 = -1$ then $\sqrt{-1} = \sqrt{i^2} \ne |i| = 1$. That $\sqrt{x^2} = |x|$ is not a definition but a consequence. And it is only true if $x$ is a real number. – fleablood Jan 19 '24 at 16:46

3 Answers3

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Squaring both side is not always reversible i.e. $$a=b\implies a^2=b^2\\\text{But, }\ \\ a^2=b^2 \;\not\!\!\!\implies a=b$$

Squaring both sides of an equation can introduce extraneous solutions. Which doesnot satisfy the original equation. We should always check the sloution whether it staisfy or not.

In the case of the given equation, $\sqrt{\frac{25}{16}}\ne\frac{-5}{4}$ So, $x=\frac{25}{16}$ is extraneous solution.

You easily say, the equation has no solution, by looking the LHS and RHS as mentioned in the comment.

I encourage you to see: Why can't you square both sides of an equation?

O M
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The square root function is the inverse of the right end of the graph $f(x)={x}^{2}$.

${f}^{-1}(x)=\sqrt{x}$ does not include negative value as its range.

Account
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To counter your example,

$$\begin{align}x^2 = 16 &\iff x^2 - (4)^2 = 0\\& \iff(x+4)(x-4) = 0\\& \iff x = 4\ \rm or \ -4.\end{align}$$

Taking square root here is not a good idea as you might lose a solution in hasle.

Further, you can't get negative value as output for the square root function. If you are a friend with graphs, have a look at the following graph. (This answer provides an explanation for the graph.)

Taking square roots both sides, or squaring both sides should be done with due care so as to make sure that you are not losing any solution or getting any absurd extraneous solution.

As mentioned in the comments, LHS is positive whereas RHS is negative, so the given equation has no solution as $\sqrt{x}$ can't be equal to a negative number.

Utkarsh
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