Question

a^2+c^2-4a-4c-ac+16=0,then what is value of(a+c)^2?

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

GIVEN

$\sf{ {a}^{2} + {c}^{2} - 4 a- 4c - ac+ 16 \: = 0 \: }$

TO DETERMINE

$\sf{ {(a + c)}^{2} \: }$

CALCULATION

$\sf{ {a}^{2} + {c}^{2} - 4 a- 4c - ac+ 16 \: = 0 \: }$

Multiplying both sides by 2 we get

$\sf{ 2{a}^{2} + 2 {c}^{2} - 8a- 8c - 2ac+ 32 = 0 \: \: }$

$\implies \: \sf{ ( {a}^{2} - 2ac + {c}^{2} ) + ( {a}^{2} + {c}^{2} - 8 a- 8c - 2ac+ 16 \: ) = 0 \: }$

$\implies \: \sf{ {(a - c)}^{2} + \bigg[ {a}^{2} - 8a + 16 \bigg] +\bigg[ {c}^{2} - 8c + 16 \bigg] = 0 \: }$

$\implies \: \sf{ {(a - c)}^{2} + {(a - 4)}^{2} +{(c - 4)}^{2} = 0 \: }$

Now if sum of the squares of three Real Numbers are zero then they are separately zero

Therefore

$\sf{a - c = 0 \: \: \: and \: \: a - 4 = 0 \: \: \: \: \: and \: \: \: c - 4 = 0 \: \: }$

Now

$\sf{a - c = 0 \: \: \: } \: \: gives \: \: \: a = c$

$\sf{a - 4 = 0 \: \: \: } \: \: \: gives \: \: \: a = 4$

$\sf{ c - 4 = 0 \: \: } \: \: gives \: \: \: \: c = 4$

Above three together gives

$\sf{a = c = 4 \: \: }$

Hence

$\sf{ {(a + c)}^{2} \: }$

$= \sf{ {(4 + 4)}^{2} \: }$

$= \sf{ {(8)}^{2} \: }$

$= \sf{ 64\: }$

RESULT

$\boxed{ \: \: \sf{ {(a + c)}^{2} \: } = 64 \: \: \: }$

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