# Find the range of values which has no real solutions

I would like to know how to solve the following problem:

Find the range of values of the parameter $$m$$ for which the equation $$2x^2 - mx + m = 0$$ has no real solutions.

I know I have to use the quadratic formula and the response is $$0 < m < 8$$. But what I don't know is how to proceed to find this answer. Thanks for your help.

No, you don't have to use the quadratic formula. Since\begin{align}2x^2-mx+m&=2\left(x-\frac m4\right)^2+m-\frac{m^2}8\\&=2\left(x-\frac m4\right)^2+\frac{8m-m^2}8\end{align}it s clear that your equation has no roots if and only if $$8m-m^2>0$$. And, since $$8m-m^2=m(8-m)$$, this occurs if and only if $$m\in(0,8)$$.

• If you multiply first by $8=4\times 2$ (four times the coefficient of $x^2$) it has the effect of clearing fractions. But this procedure of simply competing the square is the simplest and also makes it easier to see what is going on. – Mark Bennet Nov 18 '18 at 16:14

Guide:

• A quadratic equality has no real solution if and only the discriminant is negative.
• First, find the discriminant, find out when is it negative.

If you rearange equation like this $$m= {2x^2\over x-1}$$ you must figer out for witch $$m$$ graph of $$f(x) = {2x^2\over x-1}$$ does not cuts the line $$y=m$$ ?

Hint: A quadratic equation has no real roots iff the discriminant is negative.

$$\Delta = b^2-4ac$$

$$\Delta < 0 \implies b^2-4ac < 0$$

The given quadratic equation is $$\color{blue}{2}x^2\color{purple}{-m}x\color{green}{+m} = 0$$

Identifying what $$a$$, $$b$$, and $$c$$ are, you only have to set $$b^2-4ac < 0$$ and see what values of $$m$$ satisfy that condition.