#### Answer

$-5$

#### Work Step by Step

RECALL:
(i) $(a-b)(a+b) = a^2-b^2$
(ii) $\sqrt{a}^2=a, a \ge0$
Use rule (i) above to obtain:
$=1^2-\sqrt{6}^2
\\=1-\sqrt{6}^2$
Use rule (ii) above to obtain:
$=1-6
\\=-5$

Published by
Pearson

ISBN 10:
0-13417-805-X

ISBN 13:
978-0-13417-805-9

$-5$

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