Question

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Answer 1

What is the value of \[b^2 - 4ac\] for the following equation?
\[2x^2 + 3x = -1\]

Answer 2

Mya snwer is \[\sqrt{17}\]

Answer 3

Lol **my answer ._.

Answer 4

@AloneS do you know how to complete the square?

Answer 5

The square
\[\sqrt{3^2 -4(2)(-1)}\]

Answer 6

This is how you complete the square

first multiply everything by 1/a in this case a = 2

\[\frac{ 1 }{ 2 }(2x^{2}+3x = -1) = x^{2}+\frac{ 3 }{ 2 }x = -\frac{ 1 }{ 2 }\]

b = (3/2)

\[(\frac{ 3 }{ 4 })^{2} =\frac{ 9 }{16 }\]

\[x^{2}+\frac{ 3 }{ 2}+(\frac{ 9 }{ 16 }) = -\frac{ 1 }{ 2 }+\frac{ 9 }{ 16 } = \frac{ 1 }{ 16 }\]

\[x^{2}+\frac{ 3 }{ 2 }x+(\frac{ 9 }{ 16 }) = \frac{ 1 }{ 16 }\]

\[(x+\frac{ 3 }{ 4 })^{2} = \frac{ 1 }{ 16 }\]

\[(x+\frac{ 3 }{ 4 }) = \pm \frac{ 1 }{ 4 }\]

\[\frac{ -3-1 }{ 4 } = -1, \frac{ -3+1 }{ 4 } = \frac{ -1 }{ 2 }\]

Answer 7

o.m.g photon thank you<3