Solving steps for equation with exponent and addition (x^2 + x) = 2y

Question

I have this equation:

($x^2$ + x) = 2y

Which I know solves to:

x = (-1 + sqrt(1 + 8y)) / 2

x = (-1 - sqrt(1 + 8y)) / 2

However, I have no idea about the steps to reach the solved equation, any help will be greatly appreciated.

Answer 1

If you are solving for $x$ you just treat $2y$ as though it were a constant.

$x^2 + x = 2y$ so $x^2 + x - 2y = 0$.

Either use quadratic equation or complete the square.

1) Quadratic equation.

If $a = 1; b = 1; c = -2y$ ($2y$ is treated as a constant) then $ax^2 + bx + c = 0$ and the solution is

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

$\frac {-1 \pm \sqrt {1^2 - 4(1)(-2y)}}{2*1} =$

$\frac {-1 \pm \sqrt {1 +8y}}{2} $.

That's it.

2) Completing the square.

$x^2 + x = 2y$

$x^2 + 2*\frac 12 x = 2y$

$x^2 + 2*\frac 12 x + (\frac 12)^2 = 2y + (\frac 12)^2$

$(x + \frac 12)^2 = 2y + \frac 14$

$x+\frac 12 = \pm \sqrt {2y+ \frac 14}$

$x + \frac 12 = \pm \sqrt{\frac {8y + 1} 4}$

$x + \frac 12 = \pm \frac {\sqrt{1 + 8y}}2$

$x = -\frac 12 \pm \frac {\sqrt{1 + 8y}}2$.

$= \frac {-1 \pm \sqrt{1+8y}}2$.

That's it.

Answer 2

You will need the quadratic formula for $$x^2+px+q=0$$ that is $$x_{1,2}=-\frac{p}{2}\pm\sqrt{\left(\frac{p}{2}\right)^2-q}$$

Answer 3

To solve a quadratic equation such as $ax^2+bx+c=0$ for $x$ (with $a\ne0$), you need to complete the square: $$a(x+b/2a)^2=-c+b^2/2a$$ With a little further algebra you easily get the standard formula for the solutions.