Question

How do you solve \[\sqrt{5{{x}^{2}}-40}=0\]?

Answer 1

Hint: From the question, we have been asked to solve \[\sqrt{5{{x}^{2}}-40}=0\].We can solve the given equation by using some simple transformations to the given equation in the question and then simplifying it further. Now, as of process, we have to make some transformations to the given equation in the question to get it solved.

Complete step-by-step solution:
From the question, we have been given that \[\sqrt{5{{x}^{2}}-40}=0\]
Now, we have to do squaring on both sides of the equation.
By squaring on both sides of the equation, we get
\[\sqrt{5{{x}^{2}}-40}=0\]
\[\Rightarrow 5{{x}^{2}}-40=0\]
Now, shift \[-40\] from the left hand side of the equation to the right hand side of the equation.
By shifting \[-40\] from left hand side of the equation to the right hand side of the equation, we get \[5{{x}^{2}}=40\]
Now, divide both sides of the equation with \[5\].
By dividing the both sides of the equation with \[5\], we get
\[\dfrac{5{{x}^{2}}}{5}=\dfrac{40}{5}\]
\[\Rightarrow {{x}^{2}}=8\]
Apply square root on both sides of the equation.
By applying the square root on both sides of the equation, we get
\[x=\pm \sqrt{8}\]
\[\Rightarrow x=\pm \sqrt{2\left( 4 \right)}\]
We know that \[4\] is a perfect square number.
Therefore, on furthermore simplification of the equation, we get \[x=\pm 2\sqrt{2}\]
Therefore, we got the value of \[x=\pm 2\sqrt{2}\].
Both these solutions are valid, as they satisfy the original equation.
Hence, the given equation from the question is solved.

Note: We should be very careful while doing the calculation at the end of the process. We should use exact transformations that will make the given equation more simplified. Also, we should be very careful while doing the transformations and making the given equation more simplified. Also, we should be very careful while applying the square root on both sides of the equation. This is a very simple question and its calculation can be done very easily. Similarly we can solve \[\sqrt{4{{x}^{2}}-20}=0\Rightarrow 4{{x}^{2}}=20\Rightarrow {{x}^{2}}=5\Rightarrow x=\pm \sqrt{5}\].