# The set of all values of m for which $mx^2 – 6mx + 5m + 1 > 0$ for all real x is

The set of all values of m for which $$mx^2 – 6mx + 5m + 1 > 0$$ for all real x is? The answer given is $$0<=m<1/4$$

My working: $$D>=0$$

$$=> (-6m)^2 -4(m)(5m+1)>=0$$

$$=> m(4m-1)>=0$$

=> Either $$m>=1/4$$ or $$m<=0$$

Where am I going wrong?

• The discriminant should be negative not positive – DINEDINE May 26 at 7:41

You need $$mx^2-6mx+5m+1>0$$ for all real $$x$$. So, $$mx^2-6mx+5m+1$$ is not zero for all real $$x$$. The discriminant should be negative. You also need $$m$$ to be positive. This gives $$0.

However, when $$m=0$$, $$mx^2-6mx+5m+1\equiv 1$$ is positive.

Therefore, $$0\le m<\dfrac14$$.

If we want to solve $$mx^2 - 6mx + 5m + 1 > 0$$ then we do not want any real roots!

That is, the discriminant should be $$\Delta<0$$, not $$\Delta\ge 0$$.

Therefore we have $$m(4m-1)<0$$ from which you can easily conclude the result.

It should have been $$D<0$$, since then the quadratic has no roots at all (and thus is always positive or negative – the result thus obtained confirms that $$m$$ is positive after all).

Option:

$$y=m(x^2-6x+5) +1>0;$$

0) $$m=0$$

1) $$m>0$$.

A parabola opening upward.

Minimum at:

$$y'=m(2x-6)=0;$$ $$x=3;$$

$$y_{\min}=m(9-18+5)+1=$$

$$-4m+1$$;

We require: $$y_{\min}= -4m+1>0$$, or $$m<1/4$$;

Combining : $$0 \le m < 1/4$$.

2) Rule out $$m<0$$ (Parabola opening downward )