If a^2+b^2+c^2=2(a-b-c)-3 then find the value of 2a-3b+4c
Malyarpita Singh , 6 Years ago
Grade 10
1 Answers
Arun
Last Activity: 6 Years ago
a^2+b^2+c^2 = 2(a-b-c)-3 a^2+b^2+c^2-2a+2b+2c+3 = 0 (a^2-2.a.1+1^2)+(b^2+2.b.1+1^2)+(c^2+2... = 0 (a-1)^2+(b+1)^2+(c+1)^2 = 0 [if the sum of three terms equals to 0, the squares are also equals to 0 because square of something never be negative...] so, (a-1)^2 = 0 .'.a = 1 (b+1)^2 = 0 .'.b = -1 (c+1)^2 = 0 .'.c = -1
Therefore:
2a - 3b + 4c = 2 + 3 -4 = 1
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