Yes. Think of y as being a function of x.
y = f(x) = x2 + 1
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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