Given that \(\sqrt{128}+\sqrt{18}-\sqrt{K} = 7\sqrt{2}\), find K,
Question 1 Report
Given that \(\sqrt{128}+\sqrt{18}-\sqrt{K} = 7\sqrt{2}\), find K,
Answer Details
We have the expression: \(\sqrt{128}+\sqrt{18}-\sqrt{K} = 7\sqrt{2}\).
We can simplify the two radicals to get:
\(\sqrt{128}=8\sqrt{2}\) and \(\sqrt{18}=3\sqrt{2}\).
Substituting these values into the expression, we get:
\(8\sqrt{2}+3\sqrt{2}-\sqrt{K} = 7\sqrt{2}\)
Combining like terms on the left side, we have:
\(11\sqrt{2}-\sqrt{K} = 7\sqrt{2}\)
Isolating the square root on the left side, we get:
\(-\sqrt{K} = -4\sqrt{2}\)
Squaring both sides, we have:
\(K = (-4\sqrt{2})^2\)
Simplifying the expression, we get:
\(K = 32\)
Therefore, the value of K is 32.
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