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.
$\endgroup$4 Answers
$\begingroup$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 $m\in(0,8)$.
$\endgroup$ 1 $\begingroup$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.
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