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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.

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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 $m\in(0,8)$.

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  • $\begingroup$ 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. $\endgroup$ – Mark Bennet Nov 18 '18 at 16:14
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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.
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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 $y=m$ ?

enter image description here

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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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