decide whether x=2 ia a solution of x² + 4x -12
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prabhakaranmolraisa:
this is wrong
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4
putting x=-2
![{x}^{2} + 4x - 12 \\ = { - 2}^{2} + 4 \times ( - 2) - 12 \\ = 4 - 8 - 12 \\ = 4 - 20 \\ = - 16 \\ {x}^{2} + 4x - 12 \\ = { - 2}^{2} + 4 \times ( - 2) - 12 \\ = 4 - 8 - 12 \\ = 4 - 20 \\ = - 16 \\](https://tex.z-dn.net/?f=+%7Bx%7D%5E%7B2%7D++%2B+4x+-+12+%5C%5C++%3D++%7B+-+2%7D%5E%7B2%7D++%2B+4+%5Ctimes+%28+-+2%29+-+12+%5C%5C++%3D+4+-+8+-+12+%5C%5C++%3D+4+-+20+%5C%5C++%3D++-+16+%5C%5C+)
since -16!= 0
hence x=2 is not a factor of the given equation
since -16!= 0
hence x=2 is not a factor of the given equation
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