Question
The positive square root of $7+\sqrt{48},$ is:
- $7+2\sqrt3$
- $7+\sqrt3$
- $2+\sqrt3$
- $3+\sqrt2$
✓
Answer
- $2+\sqrt3$
$\sqrt{7+\sqrt{48}}$
$=\sqrt{7+2\sqrt{12}}$
$=\sqrt{4+3+2\sqrt4\times\sqrt3}$
$=\sqrt{\big(\sqrt4\big)^2+\big(\sqrt3\big)^2+2\times\sqrt4\times\sqrt3}$
$=\sqrt{\big(\sqrt4+\sqrt3\big)^2}$
$=\pm\big(\sqrt4+\sqrt3\big)$
Positive value is $\sqrt4+\sqrt3=2+\sqrt3$
Hence, correct option is (c).
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