Solving radical equations

**m0nkeysensei** · November 12th 2009, 03:53 PM

$\sqrt(k+2) - \sqrt(k-3) = 1$
$\sqrt(k+2) = \sqrt(k-3) + 1$
$(\sqrt(k+2))^2 = (1 + \sqrt(k-3))^2$
$k+2=1+\sqrt(k-3)+\sqrt(k-3)+(k-3)$
$k+2=-2+k+2\sqrt(k-3)$
$0=2\sqrt(k-3)$
$0=4(k-3)$
0=4k-12
12=4k
4=k

but the answer is k=7
Where did i mess up?

**skeeter** · November 12th 2009, 03:58 PM

Originally Posted by m0nkeysensei

$\sqrt(k+2) - \sqrt(k-3) = 1$
$\sqrt(k+2) = \sqrt(k-3) + 1$
$(\sqrt(k+2))^2 = (1 + \sqrt(k-3))^2$
$k+2=1+\sqrt(k-3)+\sqrt(k-3)+(k-3)$
$k+2=-2+k+2\sqrt(k-3)$
$0=2\sqrt(k-3) \, \, \,\, \textcolor{red}{this \, step \, is \, incorrect \, \, \, \, 4 = 2\sqrt{k-3}}$
$0=4(k-3)$
0=4k-12
12=4k
4=k

but the answer is k=7
Where did i mess up?

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