Exponents/solving imaginary expressions
Write the expression as a radical
x(2/3)
3√x2
√(100)
10
√(10-x)=-8
There is no solution because there is no positive square root that can equal a negative number
the square root of -1
i
Write the expression without radicals
3√x2
x(2/3)
√(-4) * √(-9)
5i
3√(x-12)=-4
-52
Why do we need imaginary numbers?
To find the square root of negative numbers
How many real solutions does this expression have?
x2+5x+20
√(-9)
3i
√(x-15)=3
24
Simplify:
i2
√-1
i3
-1
i
-i
Solve using the quadratic formula
x2+2x+12
x=-1+(1/2)√(32)i and -1-(1/2)√(32)i
√(-81)
9i
(1 -3i)(2+4i)
14-2i
How many solutions does a cubed root have?
1
√(36) + √(-36)
6 + 6i
Solve using competing the square
(x-5)2=-12
x=√(12)i+5 and -√(12)i+5
√(-49) + √(-64)
15i
√(8-x)+5=2
This expression is equivalent to √(8-x)=-3
so it cannot be solved because a positive square root cannot equal a negative number
√(25) + √(-100)
5 + 100i