Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is basic algebra. I must be making a mistake somewhere. Where is it?

I start out with $x^2 - x = y - 1$.

$x^2 - x - (y - 1) = 0$.

Using quadratic formula, $x = \frac{x \pm \sqrt{x^2 + 4(y - 1)}}{2}$.

$x = \pm\sqrt{x^2 + 4(y - 1)}$

$x^2 = x^2 + 4(y-1)$.

$0 = 4(y -1)$

$0 = y - 1$

$y = 1$

share|cite|improve this question

What exactly are you trying to do? If you are trying to solve for x in terms of the coefficients and $y$ (which I assume is being held constant) then you only take the coefficient of the $x$ terms when you apply the quadratic formula, so it would be $$x=\frac{1\pm\sqrt{1+4(y-1)}}{2} \\x=\frac{1\pm\sqrt{4y-3}}{2}$$ which cannot be simplified any further.

share|cite|improve this answer
Yes, thank you, I put x and x^2 instead of their coefficients. Silly misapplication of quadratic formula. – user75122 Jun 9 '13 at 2:26

$x=\dfrac{1 \pm \sqrt{1+4(y-1)}}{2}$,don't put $x$ in.

share|cite|improve this answer
Yep, this question is resolved, thanks everyone – user75122 Jun 9 '13 at 2:28

If you mean for $y$ to be a constant, then $$x^2 - x = y - 1 \iff x^2 + -x + -(y-1)$$ So if you want to use the quadratic formula to "solve" for x:

$$x = \dfrac {-b\pm \sqrt{b^2 - 4ac}}{2a}$$

then you need to substitute $a = 1, b = -1, c = -(y-1)$: $$x=\dfrac{1 \pm \sqrt{1+4(y-1)}}{2}$$No $x$ terms in the right-hand side of the quadratic formula when solving for $x$!

Otherwise, I will assume you have two variables: $x, y$, which is not quadratic equation in one variable, so the quadratic formula is inappropriate. So I will assume, as your title suggests, that you might have been asked to simplify the equation, or to express it as a function of $y$, or to express it in a standard form for a parabola:

You can take your equation and write it as a function of $y$:

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

The last equation reveals nicely that your function $f(x) = y$ is a parabola.

You can "complete the square" to write the parabola in the following form: $$\begin{align} y & = x^2 - x + 1 \\ & = (x^2 - x - 1/4) + 5/4 \\ & = (x - 1/2)^2 + 5/4 \end{align}$$

This gives a parabola in the form $$\left(y - \dfrac 54\right) = \left(x - \dfrac 12\right)^2$$ which opens "up", and has its vertex at $\left( \dfrac 12, \dfrac 54\right)$:

enter image description here

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.