waecmaths question:
If y varies directly as the square root of (x +1) and y = 6 when x =3, find x when y = 9
Option A:
8
Option B:
7
Option C:
6
Option D:
5
waecmaths solution:
$\begin{align} & y\propto \sqrt{x+1} \\ & y=k\sqrt{x+1}\text{ where }k\text{ is the constant proportionality} \\ & \text{When }x=3,\text{ }y=6 \\ & 6=k\sqrt{3+1} \\ & 6=2k \\ & k=3 \\ & \text{when }y=9,\text{ }x\text{ will be} \\ & 9=3\sqrt{x+1} \\ & 3=\sqrt{x+1} \\ & \text{Square both sides} \\ & 9=x+1 \\ & x=8 \\\end{align}$
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