Question

Can anyone give answer and explanation of this question

Answer 1

Step-by-step explanation:

Given:

$2a^{2} + 9 {b}^{2} + 4 {c}^{2} - 4a - 6b + 4ac = - 5$

To find:

$a^{2} - {b}^{2} + {c}^{2}$

Solution:

$2a^{2} + 9 {b}^{2} + 4 {c}^{2} - 4a - 6b + 4ac = - 5$

${a}^{2} + 4 {c}^{2} + 4ac + {a}^{2} + 9 {b}^{2} - 4a - 6b = - 5$

$(a + 2c)^{2} +( {a}^{2} - 4a + 4 - 4) + (9 {b}^{2} - 6b + 1 - 1) = - 5$

${(a + 2c)}^{2} + ( {a}^{2} - 4a + 4) + (9 {b}^{2} - 6b + 1) - 4- 1 = - 5$

${(a + 2c)}^{2} + {(a - 2)}^{2} + {(3b - 1)}^{2} = - 5 + 5$

${(a + 2c)}^{2} + {(a - 2)}^{2} + {(3b - 1)}^{2} = 0$

We know that, if som of three positive quantities is 0, then the individual quantity kust be equal to 0.

So,

$(a + 2c) = 0 \: \: and \: \: (a - 2) = 0 \: \: and \: \: (3b - 1) = 0$

On solving these equations, we get,

$a = 2 \: \: and \: \: b = \frac{1}{3} \: \: and \: \: c = - 1$

Now,

${a}^{2} - {b}^{2} + {c}^{2} = {(2)}^{2} - { (\frac{1}{3}) }^{2} + {( - 1)}^{2}$

$= 4 - \frac{1}{9} + 1$

$= \frac{44}{9}$

Hope this will help you....!