x2 + 7x + 9 = 3 ∴ x2 + 7x + 6 = 0 ∴ (x + 6)(x + 1) = 0 ∴ x ∈ {-6, -1}
x2 + 9 = 10x x2 - 10x + 9 = 0. (x - 9)(x - 1) = 0. Therefore, x = 1 or 9.
Proper form first. X2 + Y2 = 1 Y2 = 1 - X2 Y = (+/-) sqrt(1 - X2) -------------------------- zero out the X Y = (+/-) sqrt(1 - 02) Y = 1 ----------------the radius of this circle
A quadratic equation. If you wish to solve for x, you can do so as follows: -x2 + 6x + 7 = 0 x2 - 6x - 7 = 0 (x - 7)(x + 1) = 0 x ∈ {-1, 7}
x2 + x + 1 = 0 ∴ x2 + x + 1/4 = -3/4 ∴ (x + 1/2)2 = -3/4 ∴ x + 1/2 = ± √(-3/4) ∴ x = - 1/2 ± (i√3) / 2 ∴ x = (-1 ± i√3) / 2
y = x2 + 14x + 21 a = 1, b = 14 x = -b/2a = -14/2*1 = -7
-7.5?
y = x2 + 2x + 1zeros are:0 = x2 + 2x + 10 = (x + 1)(x + 1)0 = (x + 1)2x = -1So that the graph of the function y = x + 2x + 1 touches the x-axis at x = -1.
y = x2 + x = 0 x (X + 1) = 0 x = 0 is one solution x = -1 is the other
Interpreting that function as y=x2+2x+1, the graph of this function would be a parabola that opens upward. It would be equivalent to y=(x+1)2. Its vertex would be at (-1,0) and this vertex would be the parabola's only zero.
x(x + 1)
That is not a function, although it does involve the function of addition. A function is something that is done to numbers.
(xn+2-1)/(x2-1)
x2 + 2x -6 = 0 x2 + 2x + 1 = 7 (x + 1)2 = 7 x = -1 ± √7
The vertex is at (-1,0).
y=x2+2x+1 -b -2 2a= 2= -1 = axis of symmetry is negative one.
No.