Find the value of x for which 6\(\sqrt{4x^2 + 1}\) = 13x, where x > 0

Assessment: WAEC SSCE - Further Mathematics - 2019 (Objective) Subject: Further Mathematics

Question 1 Report

Find the value of x for which 6\(\sqrt{4x^2 + 1}\) = 13x, where x > 0

Answer Details
To solve for x in the equation 6\(\sqrt{4x^2 + 1}\) = 13x, we need to isolate x on one side of the equation. First, we can simplify the left-hand side by squaring both sides of the equation: (6\(\sqrt{4x^2 + 1}\))^2 = (13x)^2 Simplifying the left-hand side, we get: 6^2 * (4x^2 + 1) = 13^2 * x^2 Simplifying further: 144x^2 + 36 = 169x^2 Subtracting 144x^2 from both sides: 36 = 25x^2 Dividing both sides by 25: x^2 = \(\frac{36}{25}\) Taking the square root of both sides: x = \(\frac{6}{5}\) Therefore, the value of x that satisfies the equation is \(\frac{6}{5}\).
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