If y varies directly s the square root of (x + 1) and y = 6 when x = 3, find x when y = 9
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Correct Answer: Option A
Explanation:
y \(\alpha\) \sqrt{x + 1}\), y = k\sqrt{x + 1}\)
6 = k\(\sqrt{3 + 1}\)
6 = k\(\sqrt{4}\)
6 = 2k
k = \(\frac{6}{2}\) = 3
y = \(\sqrt{(x + 1)}\)
9 = 3\(\sqrt{(x + 1)}\)(divide both side by 3)
\(\frac{9}{3}\) = \(\frac{3\sqrt{x + 1}}{3}\)
3 = \(\sqrt{x + 1}\)(square both sides)
9 = x + 1
x = 9 - 1
x = 8
y \(\alpha\) \sqrt{x + 1}\), y = k\sqrt{x + 1}\)
6 = k\(\sqrt{3 + 1}\)
6 = k\(\sqrt{4}\)
6 = 2k
k = \(\frac{6}{2}\) = 3
y = \(\sqrt{(x + 1)}\)
9 = 3\(\sqrt{(x + 1)}\)(divide both side by 3)
\(\frac{9}{3}\) = \(\frac{3\sqrt{x + 1}}{3}\)
3 = \(\sqrt{x + 1}\)(square both sides)
9 = x + 1
x = 9 - 1
x = 8