Question

i would be very grateful if u answer this question

Answer 1

$\blue{\bold{\underline{\underline{Answer: \: 3 }}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green { \tt{ a^2 + 9b^2 + c^2 = 2(a + 3b + 2c) - 6 }}$

$\pink{ \tt{ \leadsto a^2 + 9b^2 + c^2 = 2a + 6b + 4c - 6 }}$

$\green { \tt { \mapsto a^2 + 9b^2 + c^2 - 2a - 6b - 4c + 6 = 0 }}$

$\pink{ \tt{ \leadsto a^2 - 2a + 9b^2 - 6b + c^2 - 4c + (1 + 4 + 1) = 0 }}$

$\green { \tt{ \mapsto a^2 - 2a + 1 + 9b^2 - 6b + 1 + c^2 - 4c + 4 = 0 }}$

$\pink{ \tt{ \leadsto (a^2 - 2a + 1) + (9b^2 - 6b + 1) + (c^2 - 4c + 4) = 0 }}$

$\green { \tt{ \mapsto [ a^2 - a - a + 1 ] + [ 9b^2 - 3b - 3b + 1 ] + [ c^2 - 2c - 2c + 4 ] = 0 }}$

$\pink { \tt{ \leadsto [ a(a - 1) - 1(a-1) ] + [ 3b( 3b - 1) - 1( 3b - 1) ] + [ c(c-2) - 2(c-2) ] = 0 }}$

$\green { \tt { \mapsto [ a - 1 ]^2 + [ 3b - 1 ]^2 + [ c - 2 ]^2 = 0 }}$

$\blue{ \tt{ The \: square \: of \: any \: non \: zero \: integer \: always \: has \: to \: be \: positive }}$

( > 0)

$\purple { \tt { So , \: If \: the \: sum \: of \: the \: square \: of \: any \: three \: integral \: values \: is \: equal \: to \: 0 , \: each \: integral \: value \: is \: equal \: to \: 0 }}$

$\orange { \tt { (a-1)² = 0 . This \: gives \: a = 1 }}$

$\blue { \tt{ (3b - 1)² = 0. This \: gives \: b = ⅓ }}$

$\orange { \tt { (c-2)² = 0. This \: gives \: c = 2 }}$

$\green { \tt { Substituting \: the \: values }}$

$\sf{ \pink{ \sqrt { \dfrac{a+c}{b}} \: is \: equal \: to \: \leadsto \sqrt { \dfrac{1+2}{\dfrac{1}{3}}}}}$

$\sf{ \pink { \leadsto \sqrt{ \dfrac{3}{ \dfrac{1}{3} } } } }$

$\sf{ \pink { \leadsto \sqrt{ 9 }}}$

Answer : The required value is 3 .