Solving imaginary Equations
18 - x2 = 39
-x2 = 21
x2 = -21
x = +/- i√21
√(-9)
3i
(2+3i) + (4+7i)
What is 6+10i
√(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
x2+ 13 = 1
x = +/- 2i√3
√(-81)
9i
3√(x-12)=-4
-52
Why do we need imaginary numbers?
To find the square root of negative numbers
4x2 + 15 = -9
x = +/- i√6
8i * 4i
32i2 = -32
(6+5i)- (3+2i)
What is 3+3i
√(x-15)=3
24
Simplify:
i2
√-1
i3
-1
i
-i
Solve the quadratic equation
-3x2 = 150
x= +/- 5i√2
√(-4) * √(-9)
√(-4) * √(-9)
i√4 * i√9
2i * 3i
6i2 = 6 * (-1)
-6
(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