Show that x−x2=13 has no real solutions.

Question

Lughettitbn · Accepted Answer

I mean the easiest way is to use the discriminant (as mentioned in the first comment). But hey, why use the quadratic equation when you can just complete the square (i.e. derive the quadratic equation):
$x^{2} - x + \frac{1}{3} = 0$

$(x - \frac{1}{2})^{2} + \frac{1}{3} + ? = 0$

$x^{2} - x + \frac{1}{4} + \frac{1}{3} + ? = 0 \to ? = - \frac{1}{4}$

$(x - \frac{1}{2})^{2} + \frac{1}{3} - \frac{1}{4} = 0$

$(x - \frac{1}{2})^{2} + \frac{1}{12} = 0$

$(x - \frac{1}{2})^{2} = - \frac{1}{12}$
A real value cannot be squared (positive or negative) to give a negative result, so this has no real solutions.
If we inspect $b^{2} - 4 a c$ , we find:
$1 - \frac{4}{3} = - \frac{1}{3} < 0 \to no real solutions$
${(2 (x - \frac{1}{2}))}^{2} = {(2 x - 1)}^{2} = 4 {(x - \frac{1}{2})}^{2}$
So we can use that and just multiply each side by 4
${(2 x - 1)}^{2} = - \frac{1}{3}$

Show that x-x^2=\frac{1}{3} has no real solutions.

Answered question

Answer & Explanation

New Questions in Algebra I