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Nadya [2.5K]
3 years ago
12

Is 2i a real or complex root?

Mathematics
1 answer:
Maurinko [17]3 years ago
3 0

By definition, i = sqrt(-1) because it is used to solve the equation x^2+1 = 0 or x^2 = -1, which has no real number solutions.

Since i is imaginary or complex, then so is 2i

You can express it as 2i = 0+2i to be in the form a+bi with a = 0 as the real part and b = 2 as the imaginary component.

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We want to find out when it returns to ground level (h = 0)

To find this out, we can plug in 0 and solve for t.

0 = -16t+64t+3 \\ 16t-64t-3=0 \\ use\ the\ quadratic\ formula\ \frac{-b\±\sqrt{b^2-4ac}}{2a}  \\ \frac{-(-64)\±\sqrt{(-64)^2-4(16)(-3)}}{2*16}  = \frac{64\±\sqrt{4096+192}}{32}

= \frac{64\±\sqrt{4288}}{32} = \frac{64\±8\sqrt{67}}{32} = \frac{8\±\sqrt{67}}{4} = \boxed{\frac{8+\sqrt{67}}{4}\ or\ 2-\frac{\sqrt{67}}{4}}

So the ball will return to the ground at the positive value of \boxed{\frac{8+\sqrt{67}}{4}} seconds.

What about the vertex? Simple! Since all parabolas are symmetrical, we can just take the average between our two answers from above to find t at the vertex and then plug it in to find h!

\frac{1}2(\frac{8+\sqrt{67}}{4}+2-\frac{\sqrt{67}}{4}) = \frac{1}2(2+\frac{\sqrt{67}}{4}+2-\frac{\sqrt{67}}{4}) = \frac{1}2(4) = 2

h=-16t^2+64t+3 \\ h=-16(2)^2+64(2)+3 \\ \boxed{h=67}


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