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solution set for the equation -2x^2+8x-8=0 A) {-2} B) {2} C) {-2,2} J) {-2, 0, 5}
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D*
definitely not D
you have a polynomial of degree two (quadratic) and it can have at most 2 zeros, not 3 also notice that if you replace x by 0 you do not get 0, you get -8
do you know how i can get the answer?
you could use the quadratic formula \[x=\left( \left( -b \pm \sqrt{b ^{2}-4ac} \right)\over 2a \right)\]where\[y=ax ^{2}=bx=c\]
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No I meant \[ax ^{2}+bx+c=0\]
I guess I'm getting tired and making silly mistakes.
thank you :D
you're welcome.
is that possible with that becuase the number inside the radical is negative D:
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