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Mathematics 15 Online
OpenStudy (anonymous):

Use the quadratic formula to solve for x. 2x^2 + 3x - 7 = 0

OpenStudy (anonymous):

$$\Large \begin{array}{*{20}c} {x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} & {{\rm{when}}} & {ax^2 + bx + c = 0} \\ \end{array} $$

OpenStudy (anonymous):

You can plug in now.

OpenStudy (anonymous):

\[x=\frac{ -3\pm \sqrt{3}^{2}-4\times2\times-7 }{ 2\times2 } \] is this correct?

OpenStudy (steve816):

You need to square root the entire b^2 - 4ac

OpenStudy (anonymous):

ahh yeah I see I forgot to add it into the equation drawing. \[x=\frac{ 3\pm \sqrt{65} }{ 4 }\]

OpenStudy (anonymous):

I cannot add a \[\pm sign\] to the answer, so this last one is incorrect? I think I need to put in two answers. How can I get the two answers. Do I need to put in both \[x=\frac{ 3+\sqrt{65} }{ 4 }\] and \[x=\frac{ 3-\sqrt{65} }{ 4 }\] ? or is there another answer?

OpenStudy (loser66):

yup

OpenStudy (loser66):

you are correct

OpenStudy (anonymous):

about adding both the answers? \[x=\frac{ 3+\sqrt{65} }{ 4 }\] and \[x=\frac{ 3-\sqrt{65} }{ 4 }\] ? instead of one that has the \[\pm sign\] ?

OpenStudy (loser66):

yup, you can do either way, follow your prof's way. If he used \(\pm\) then, use it. To me, separating is ALWAYS righ, \(\pm\) is not wrong but to the picky prof, you might lose your credits because of it.

OpenStudy (anonymous):

ok ty

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