x2 + 2x - 15 = (x - 3)(x + 5)
2x = x2 + 4x - 3 x2 + 2x - 3 = 0 (x - 1)(x - 2) = 0
have to put it was 2x and not x2--back to the problem--2x-x-4=0x=4* * * * *True, but the question, in all likelihood WAS x2 - x - 4 = 0and the solutions to that equation are -1.561553 and 2.561553
x2 + 2x -6 = 0 x2 + 2x + 1 = 7 (x + 1)2 = 7 x = -1 ± √7
What do you want to convert it to? x2 + y2 = 2x If you want to solve for y: x2 + y2 = 2x ∴ y2 = 2x - x2 ∴ y = (2x - x2)1/2 If you want to solve for x: x2 + y2 = 2x ∴ x2 - 2x = -y2 ∴ x2 - 2x + 1 = 1 - y2 ∴ (x - 1)2 = 1 - y2 ∴ x - 1 = ±(1 - y2)1/2 ∴ x = 1 ± (1 - y2)1/2
x2 + 2x - 15 = (x - 3)(x + 5)
2x = x2 + 4x - 3 x2 + 2x - 3 = 0 (x - 1)(x - 2) = 0
x2-2x-15 = (x+3)(x-5) when factored
have to put it was 2x and not x2--back to the problem--2x-x-4=0x=4* * * * *True, but the question, in all likelihood WAS x2 - x - 4 = 0and the solutions to that equation are -1.561553 and 2.561553
x2 + 2x -6 = 0 x2 + 2x + 1 = 7 (x + 1)2 = 7 x = -1 ± √7
x2 - 2x - 5 = 0 x2 - 2x + 1 = 6 (x - 1)2 = 6 x - 1 = ± √6 x = 1 ± √6
What do you want to convert it to? x2 + y2 = 2x If you want to solve for y: x2 + y2 = 2x ∴ y2 = 2x - x2 ∴ y = (2x - x2)1/2 If you want to solve for x: x2 + y2 = 2x ∴ x2 - 2x = -y2 ∴ x2 - 2x + 1 = 1 - y2 ∴ (x - 1)2 = 1 - y2 ∴ x - 1 = ±(1 - y2)1/2 ∴ x = 1 ± (1 - y2)1/2
f(x)=cos(sin(x2)) [u(v)]' = u'(v) * v' so f'(x) = cos'(sinx(x2)) * sin'(x2) * (x2)' f'(x) = -sin(sin(x2)) * cos(x2) * 2x = -2x sin(sin(x2)) cos(x2)
x2 - 2x - 13 = 0 x2 - 2x + 1 = 14 (x - 1)2 = 14 x - 1 = ±√14 x = 1 ±√14 x ≈ {4.742, -2.742}
x2 - 10x - 6 = 2x + 1 Subtracting 2x + 1 from both sides gives: x2 - 12x - 7 = 0
x2-2x-3 = 0 (x+1)(x-3) = 0 x = -1 or x = 3
(2x)ysquared