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Mathematics
OpenStudy (anonymous):

Can someone please help me with this question? I have tried working it out several times and have gotten it wrong every time.... Here is the question...Solve t^2 + 4t = 1 quadratic formula ax^2 + bx + c = 0 t^2 + 4t – 1 = 0

OpenStudy (anonymous):

\[t^2+4t=1\] \[(t+2)^2+1+2^2=5\] \[t+2=\pm\sqrt{5}\] \[t=-2\pm\sqrt{5}\]

OpenStudy (anonymous):

or use \[t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] with \[a=1,b=4,c=-1\] and get \[t=\frac{-4\pm\sqrt{4^2-4\times 1\times (-1)}}{2\times 1}\] \[t=\frac{-4\pm\sqrt{16+4}}{2}\] \[t=\frac{-4\pm\sqrt{20}}{2}\]

jimthompson5910 (jim_thompson5910):

The answers are \[\large t=-2+\sqrt{5} \textrm{ or } t=-2−\sqrt{5}\] See the attached image for how I got those answers using the quadratic formula.

OpenStudy (anonymous):

but then you have to "simplify" this because \[\sqrt{20}=\sqrt{4\times 5}=2\sqrt{5}\] so you have \[\frac{-4\pm\2\sqrt{5}}{2}=\frac{2(-2\pm\sqrt{5})}{2}=-2\pm\sqrt{5}\]

OpenStudy (anonymous):

Thank you so much... I have been stuck on this one question... I have been trying to do homework and do my final exam for today...

OpenStudy (anonymous):

same answer, but it is a lot easier to complete the square when the coefficient of the "middle term" is even, because then you do not have to spend all that time simplifying the radical a the end

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