x2 + 4x + 3 = 0 (x + 1)(x + 3) = 0 x ∈ {-3, -1}
x2 + 13x = -30 ∴ x2 + 13x + 30 = 0 ∴ (x + 3)(x + 10) = 0 ∴ x ∈ {-3, -10}
x2+6x+9 = x+3 x2+6x-x+9-3 = 0 x2+5x+6 = 0 Solve by factoring or with the help of the quadratic equation formula: (x+3)(x+2) = 0 Therefore: x = -3 or x = -2
If we assume that it equals zero and you wish to solve for x, then the answer is:-x2 + 2x - 3 = 0x2 - 2x + 3 = 0x2 -2x + 1 = -2(x - 1)2 = -2x - 1 = ± √(-2)x = 1 ± i√2
x2 + 10x + 21 = (x + 3)(x + 7)
x2 plus 3x plus 3 is equals to 0 can be solved by getting the roots.
x2 + 4x + 3 = 0 (x + 1)(x + 3) = 0 x ∈ {-3, -1}
x2 + 13x = -30 ∴ x2 + 13x + 30 = 0 ∴ (x + 3)(x + 10) = 0 ∴ x ∈ {-3, -10}
x2 + 3y = 7 3x + y2 = 3 3y = x2 + 7 y2 = -3x + 3 y = x2/3 + 7/3 y = ± √(-3x + 3) If you draw the graphs of y = x2/3 + 7/3 and y = ± √(-3x + 3) in a graphing calculator, you will see that they don't intersect, so that the system of the given equations has not a solution.
x2+6x+9 = x+3 x2+6x-x+9-3 = 0 x2+5x+6 = 0 Solve by factoring or with the help of the quadratic equation formula: (x+3)(x+2) = 0 Therefore: x = -3 or x = -2
I suggest you use the quadratic formula. In this case, a = 1, b = 5, c = 3.
If we assume that it equals zero and you wish to solve for x, then the answer is:-x2 + 2x - 3 = 0x2 - 2x + 3 = 0x2 -2x + 1 = -2(x - 1)2 = -2x - 1 = ± √(-2)x = 1 ± i√2
x2 + 10x + 21 = (x + 3)(x + 7)
x2 + 2x - 13 = 2 x2 + 2x - 15 = 0 (x + 5)(x - 3) = 0 x + 5 = 0 and x - 3 = 0 so x = -5 and x = 3
In general, no.
x = 7/2 and also x = -3/2 by using the quadratic equation formula.
y = 4x-3