Question

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

Answer 1

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

Answer 2

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

Answer 3

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.

Answer 4

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

Answer 5

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...

Answer 6

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