Krdytkyu

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. Sincebegin{align}2x^2-mx+m&=2left(x-frac m4right)^2+m-frac{m^2}8\&=2left(x-frac m4right)^2+frac{8m-m^2}8end{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 $min(0,8)$.

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 $$m= {2x^2over x-1}$$ you must figer out for witch $m$ graph of $f(x) = {2x^2over x-1}$ 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 $$color{blue}{2}x^2color{purple}{-m}xcolor{green}{+m} = 0$$

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.