Question

(a-1)^2 + (b+2)^2+(c+1)^2=0, find 2a-3b+7c

Answer 1

$Given \: (a-1)^{2} + (b+2)^{2} + (c+1)^{2} = 0$

$\boxed { \pink { If \:x^{2} + y^{2}+ z^{2} = 0 \:then \: x = y = z = 0 }}$

$Here , a - 1 = 0 , \: b + 2 = 0 \: and \: c + 1 = 0$

$a = 1 , \: b = -2 , \: and \: c = -1$

$Value \: of \: 2a - 3b + 7c \\= 2\times 1 - 3\times (-2) + 7 \times (-1) \\= 2 + 6 - 7 \\= 8 - 7 \\= 1$

Therefore.,

$\red {Value \: of \: 2a - 3b + 7c} \green {= 1}$

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Answer 2

Step-by-step explanation:

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