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

Question

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?

Answer 1

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 $0<m<\dfrac 14$.

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

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

Answer 2

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.

Answer 3

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).

Answer 4

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 )