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Solve the equation if a²+b²+c²=2(a-b-c)-3 then find out the value of 2a-3b+4c=?


4 years ago

jagdish singh singh
173 Points
							$\hspace{-0.7 cm} we can write a^2+b^2+c^2=2(a-b-c)-3\\\\ as a^2-2a+1+b^2+2b+1+c^2+2c+1=0\\\\ So (a-1)^2+(b+1)^2+(c-1)^2=0\\\\ Using sum of square of all three real quantity is zero, if all are zero\\\\ So a-1 = 0 and b+1 = 0 and c+1=0\\\\ So a=1\;\;,b=-1\;\;,c=-1. So 2a-3b+4c =2+3-4=1$

4 years ago
ankit singh
596 Points
							a² + b² + c² = 2 (a - b - c ) - 3=> a² + b² + c² = 2a - 2b - 2c -3=> a² + b² + c² - 2a + 2b + 2c + 3 = 0//rearrange the above as below=> a² - 2a + 1 + b² + 2b + 1  + c² + 2c + 1 = 0=> (a-1)² + (b+1)² + (c+1)² = 0Using sum of all three real quantity is zero, if all are zero=>  a - 1  = 0 => a = 1     b + 1  = 0  => b = -1     c + 1 = 0 => c = - 1Thus value of 2a - 3b + 4c = 2 * 1 - 3(-1) + 4(-1) = 1

3 months ago
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