Question 65

If $$9a^2 + 16b^2 + c^2 + 25 = 24 (a + b)$$, then the value of (3a + 4b + 5c) is:

Solution

$$9a^2 + 16b^2 + c^2 + 25 = 24 (a + b)$$

$$9a^2+16b^2+c^2+25-24a-24b=0$$

$$9a^2+16-24a+16b^2+9-24b+c^2=0$$

$$\left(3a-4\right)^2+\left(4b-3\right)^2+c^2=0$$

So, 3a = 4, 4b=3, c=0

(3a + 4b + 5c) = 4+3+0= 7.

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