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WAEC SSCE - General Mathematics - 2004 (Objective)

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.
Read lesson note on Positive And Negative Integers, Rational Numbers (WAEC)
Read lesson note on Surds (radicals) (WAEC)
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