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 \(\frac{6}{5}\).