Y=x^2 - 2x : how do I solve for (x)?

Question

y = x^2 - 2x

how do I get (x) to be the subject?

please show steps.

Answer 1

Since this is a quadratic equations, there will be 2 solutions for x, either

1. Two distinct real values of x

2. Two identical real values of x

3. Two imaginary values of x

Rewrite the above function as follows.

x² - 2x - y = 0

Using the general solution for solving quadratic equations,

x = (2 ± √(4 + 4y))/2

= 1 ± √(1 + y)

Depending on whether y > -1, y = -1 or y < -1, you'll get the solutions of x of (1), (2) and (3) mentioned above.

If you simply want to make x the subject to do something else (eg differentiation), there are ways to do it without having to make x the subject.

Answer 2

You have to first complete the square to make it look like a quadratic equation.  The last term is equal to (b/2)^2.

So the last term for this problem would be (-2/2)^2 = (-1)^2 = 1.

Since you're adding one to side of the equation, you must do the same to the other side of the equation:

y + 1 = x^2 - 2x + 1.  x^2 - 2x + 1 can be simplified to (x -1)(x-1).

y + 1 = (x - 1)(x - 1)

y + 1 = (x - 1)^2.  Take the square root of both sides:

sqrt(y + 1) = x - 1.  Add one to both sides to isolate x:

1 + sqrt(y + 1) = x

And you're done!

Answer 3

y=x(x-2)

y/x=x-2

y/x+2=x

y+2=x^2

x=sq. root of y+2

somethin like that its been a while

Answer 4

This is x^2 - 2x = y.

Use completing the square, or the quadratic formula.

x^2 - 2x + 1 = y + 1

(x-1)^2 = y+1

x-1 = ±√(y+1)

x = 1 ± √(y+1)

Answer 5

If ax² + bx + c = 0 then x = {-b +/- sqrt(b^2 - 4ac) }/2a

this eqn is of form

x^2 -2x -y =0