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Question 13
If $$a^2 + b^2 + c^2 + 96 = 8(a + b - 2c)$$, then $$\sqrt{ab - bc + ca}$$ is equal to:
Solution
$$a^2 + b^2 + c^2 + 96 = 8(a + b - 2c)$$
$$a^2 + b^2 + c^2 + 96 - 8a - 8b + 16c = 0$$
$$(a^2 - 8a + 16) +Â (b^2 - 8b + 16) + (c^2 + 16c + 64) = 0$$
$$(a - 4)^2 + (b - 4)^2 + (c + 8)^2 = 0$$
$$(a - 4)^2 = 0, (b - 4)^2 = 0, (c + 8)^2 = 0$$
a = 4,
b = 4,
c = -8
$$\sqrt{ab - bc + ca}$$
=Â $$\sqrt{4 \times 4 + 4 \times 8 - 4 \times 8 }$$
= $$\sqrt{16 + 32 - 32 }$$
= 4
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