how do you do the square root of a negative number that cannot be perfectly squared such as -8 or -50?

Question

please show how to solve with steps
_ __
a.(-2+\ / -8) + (5- \ / -50)


( \ / is square root)

Answer 1

√(- 8) = √ (8 i ²) = √[( 4 )( 2 )( i ² )] = 2√2 i
√(- 50) = √[ ( 25 )( 2 )( i ² ) ] = 5√2 i

Answer 2

In the set of real numbers, the square root of a negative number does not exist. However in the set of complex numbers a special " number " is defined such as ( i )^2 = -1 Now the square root of a negative number is possible. For example since ( 3i )^2 = 9 i^2 and i^2 = - 1 we get ( 3i )^2 = -9 .Then the square root of -9 is 3i.

Answer 3

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RE:
how do you do the square root of a negative number that cannot be perfectly squared such as -8 or -50?
please show how to solve with steps
_ __
a.(-2+\ / -8) + (5- \ / -50)


( \ / is square root)

Answer 4

Como has it right. We are in the realm of complex numbers, so there is no point looking at your calculator.
\/[-8] = {+/-}(2\/2)i, and \/[-50] = {+/-}(5\/2)i, so there are additional hidden difficulties.

The next piece of complex number algebra needs more clarity from the questioner ( for example, does the symbol a indicate a real number?).

But mathematicians do like to keep the number \/2, it is exact.

Answer 5

ok this is a simple problem

you show a negative square root as i. so \ / -8 would be \ / 8 i
so it would simplify to 2 \ / 2 i

(-2 + 2 \ / 2 i) + (5 - 5 \ / 2 i) Now you can add
3 - 3 \ / 2 i simplify
= 3 ( 1 - \ / 2 i )
P.S. \ / is square root sign

Answer 6

I can't read the question--it's not displaying properly. What are those bars over the print?

Most people use a caret ^ to express exponents. The square root of -8 would be written (-8)^(0.5).

Answer 7

square root of negative one is called imaginary number one denoted by an i
square root of -8 is square root of -1 x square root of 8
square root of 8 is 2 square root 2
so 2 sqaure roo 2 i

Answer 8

I believe the square root of -1 is "i" so the problem should work out to -2+2i\ /2+5-5i\ /2 which comes out to 3-3i\ /2.

\ /-8= \ /-1x2x4 -1 comes out as i and 4 comes out as 2 which leaves the other 2 inside the square root symbol thus you get 2i\/2

\ /-50=\ /-1x25x2 same as above. -1 comes out as i, 25 comes out as 5 and the 2 remains under the symbol. thus you have 5i\ /2

Answer 9

-2 +sqrt(-8) + 5 - sqrt (-50)
add -2 and 5,
3 + sqrt(-8) - sqrt(-50)

Since sqrt(-8) = sqrt ((4)(-2)) = sqrt(4) x sqrt(-2) = 2 sqrt(-2)
Also sqrt (-1) = i <<<< imaginary number
sqrt(-8) = 2 sqrt(2) sqrt(-1) = 2isqrt(2)

Similarly for sqrt(-50) = sqrt(25) x sqrt(2) x sqrt(-1)
= 5isqrt(2)

Adding all the terms above, grouping like terms:
3 + 2isqrt(2) - 5isqrt(2) = 3 + isqrt(2)(2 - 5)
= 3 - 3isqrt(2) << answer

Edit: This answer agrees with the 1st answer, the one generated by computer, and the one below me (but it's not in a simplified form, sqrt(-18) =sqrt((9)(2)(-1) = 3isqrt(2))

Answer 10

I'm using rt. to mean square root.and i to indicate rt(-1)
(-2+rt(-8))=-2+2irt2
(5-2rt(-50))=5-rt(-50)=5-irt(25X2)=5-5irt(2)

So adding we get(keeping real and complex apart)

real =5-2=3 ; complex =2irt(2)-5irt(2)

ending with 3-3irt(2)

Or equivalently factorising the above(but no need to)

3(1-1rt(2))