what is the square root of -60? What is the square root of -160?

Answer 1

imaginary number solution

√- 60 =

i√60 =

√4 i√15 =

2 i √15

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√- 160 =

i√160 =

√16 i √10 =

4 i √10

- - - - - - - - -s-

Answer 2

If we're only dealing with real numbers (numbers found on the real number line), then there is no solution, as there is no real number which when multiplied by itself can be a negative number.

If we're extending the possibilities to the complex number system (and if you haven't learned about imaginary numbers or complex numbers in class, then ignore all of the following), then using the formal definiton i = √-1, we have:
√-60 = √(-1*60) = √-1 √60 = i√60 = i√(4*15) = 2i√15

√-160 = i√160 = i√(16*10) = 4i√10

Answer 3

These are 'imaginary numbers'. You also posted a question about the quadratic formula. I assume you're plotting quadratic equations (parabolas).

The roots of a parabola are the points where the curve is "anchored" to the X axis. If the quadratic gives real roots, that is, the sqrt of positive numbers, then the graph crosses the X axis at those (or that) point.

If the quadratic formula gives you imaginary numbers, the graph has no "real roots". It does not cross the X axis. The curve begins above or below the X axis and continues to increase or decrease.

Hope that helps.

Answer 4

√-60 = 2i√15

√-160 = 4i√10

Answer 5

You can't have a square root of a negative number. It would become an imaginary number. There I can't remember.

Answer 6

(-60)^(1/2) = ((-1)(4)(15))^(1/2) = 2i*(15^(1/2))

(-160)^(1/2) = ((-1)(16)(10))^(1/2) = 4i*(10^(1/2))

i = (-1)^(1/2)

Answer 7

excel is easiest, just look on top for sqaure root symbol, enter all #'s in 1 collum and go!

Answer 8

sqrt(-60)=2i*sqrt(15)

sqrt(-160) = 4i*sqrt(10)

Answer 9

sqrt(-60) = 7.746 i
sqrt(-160) = 12.649 i

where i = sqrt(-1)