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Algebra and Applications Test 3

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Question : 18 of 25

Marks: +1, -0

If

\sqrt{16-6\sqrt{7}}=a+b\sqrt{7}

what can be the value of b?

-2
$-\frac{3}{2}$
-1
$-\frac{1}{2}$
None of these

Solution:

Concept:

The radical can be simplified by writing the radicand as a perfect square.

Explanation:

16-6\sqrt{7}=9+7-6\sqrt{7}

=(3)^2+(\sqrt{7})^2-2(3)(\sqrt{7})

=(3-\sqrt{7})^2

Since

3>\sqrt{7}

,

\sqrt{(3-\sqrt{7})^2}=3-\sqrt{7}

So,

3-\sqrt{7}=a+b\sqrt{7}

Comparing terms gives

a=3

and

b=-1

Answer:

b=-1

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