x2 + x + 20 = 0; a = 1, b = 1, and c = 20; use the quadratic formula
x = [-b ± √(b2 - 4ac)]/(2a)
x = [-1 ± √(12 - 4*1*20)]/(2*1)
x = (-1 ± √-79)]/2
x = (-1 ± i√79)]/2
The solution set: {-1 - i√79, -1 + i√79}.
x2 - 4x = 8 x2 - 4x + 4 = 8 + 4 (x - 2)2 = 12 x - 2 = +&- sq. root of 12 x = 2 +&- 2(sq. root of 3)
3
x2 - x = 20 x = 5
x2 + x2 = 2x2
Substitute the value for y given by the second equation into the first equation to result in 3x - 15 = x2 - x -20. Subtract (3x - 15) from both sides and exchange sides to yield x2 - 4x - 5 = 0. This can be factored into (x - 5)(x + 1) = 0, which is true when x equals either 5 or -1. If x = 5, y = 0 and if x = -1, then y = -18. Therefore, either 5,0 or -1,-18 is an ordered pair solution to the given equation system.
x = { +6, -6 }
x2 + 49 = 0 ∴ x2 = -49 ∴ x = 7i
Using the quadratic formula, I found the solution set is x=2,x=-9
x2≤64
-4 and 4
x2=20x 0 is an obvious solution if x is not 0, then divide both side by x, so x=20 so the solutions are 0 and 20
-4
x2=9x-20 x2-9x+20=0 Factor: (x-4)(x-5)=0 x={4,5}
(-4,6)
do it yourself
There is no one solution to this problem. For example: y = x2 y = 5x y = x + 20 y = 2x / 5 + 23 These are just four of the infinite set functions that would give "y" a value of 25 when x = 5.
x2 - 4x = 8 x2 - 4x + 4 = 8 + 4 (x - 2)2 = 12 x - 2 = +&- sq. root of 12 x = 2 +&- 2(sq. root of 3)