JAMB 1993 · UME · Q10

Quadratics & their graphs

If $\sqrt{x^2 + 9} = x + 1$ , solve for $x$ .

Worked solution (try it first)

Square both sides: $x^2 + 9 = (x + 1)^2 = x^2 + 2x + 1$ .
The $x^2$ terms cancel: $9 = 2x + 1$ , so $2x = 8$ and $x = 4$ .
Check: $\sqrt{16 + 9} = 5$ and $4 + 1 = 5$ .
So $x = 4$ , option B.

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