Question 71

If $$16x^2 + 9y^2 + 4z^2 = 24(x - y + z) - 61$$, then the value of (xy + 2z) is:

Solution

As per the given question,

$$16x^2 + 9y^2 + 4z^2 = 24(x - y + z) - 61$$

$$16x^2+9y^2+4z^2-24x+24y-24z+61=0$$
$$16x^2-24x+9+9y^2+24y+16+4z^2-24z+36=0$$
$$(4x-3)^2+(3y+4)^2+(2z-6)^2=0$$
4x-3 = 0 => x = $$\dfrac{3}{4}$$

3y+4=0 => y = $$\dfrac{-4}{3}$$

2z-6 = 0 => z = 3

Hence, xy+2z = $$\dfrac{3}{4}\times\dfrac{-4}{3} + 2\times3 = -1+6 = 5$$


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