Radicals
√(32)
4√2
x2-16=0
x=4, x=-4
x2-12x+36=0
x=6
√(-36)
6i
y=-3x2
y=6x+3
(-1,-3)
√(4x2)
2x
3x2+12=0
No real solutions
2x2-x-1=0
x=1, x=-1/2
4x+2i=8+yi
x=2, y=2
y=-x+7
y=-x2-2x-1
No solutions
√(48n5)
4n2√(3n)
2x2-98=0
x=7, x=-7
9x2-6x+1=0
x=1/3
9x-18i=36+6yi
x=-4, y=-3
y=2x2+3x-4
y-4x=2
(2,10),(-3/2,-4)
√(4/49)
2/7
9x2-35=14
x=7/3, x=-7/3
-3x2+6x=4
No real solutions
(8+3i)2
55+48i
y=-2x2+x-3
y=2x-2
No solutions
√(a3/49)
a√(a)/7
(4x+5)2=9
x=-1/2, x=-2
2x2+9x+7=3
x=-1/2, x=-4
Find a possible pair of integer values for a and c so that the quadratic equation -4x+c=-ax2 has two imaginary solutions.
ac>16
y=2x-1
y=x2
(1,1)