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Question

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

How do you solve 5x240=0\sqrt{5{{x}^{2}}-40}=0?

Explanation

Solution

From the question, we have been asked to solve 5x240=0\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 5x240=0\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
5x240=0\sqrt{5{{x}^{2}}-40}=0
5x240=0\Rightarrow 5{{x}^{2}}-40=0
Now, shift 40-40 from the left hand side of the equation to the right hand side of the equation.
By shifting 40-40 from left hand side of the equation to the right hand side of the equation, we get 5x2=405{{x}^{2}}=40
Now, divide both sides of the equation with 55.
By dividing the both sides of the equation with 55, we get
5x25=405\dfrac{5{{x}^{2}}}{5}=\dfrac{40}{5}
x2=8\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=±8x=\pm \sqrt{8}
x=±2(4)\Rightarrow x=\pm \sqrt{2\left( 4 \right)}
We know that 44 is a perfect square number.
Therefore, on furthermore simplification of the equation, we get x=±22x=\pm 2\sqrt{2}
Therefore, we got the value of x=±22x=\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 4x220=04x2=20x2=5x=±5\sqrt{4{{x}^{2}}-20}=0\Rightarrow 4{{x}^{2}}=20\Rightarrow {{x}^{2}}=5\Rightarrow x=\pm \sqrt{5}.