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OpenStudy (ayee_ciera):
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OpenStudy (ayee_ciera):
what are the solutions to the quadratic equation of \[2x ^{^{2}}-16x+32=0\]and how would i solve this?
OpenStudy (dayakar):
\[2x^{2}-8x-8x+32\]=0
OpenStudy (dayakar):
2x(x-4)-8(x-4)=0
OpenStudy (dayakar):
(x-4)(2x-8)=0
OpenStudy (dayakar):
x-4=0 or 2x-8 =0 x=4 or x=8/2 =4 x=4 that means equal roots
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OpenStudy (dayakar):
are u following
OpenStudy (ayee_ciera):
yes @dayakar
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