Quadratic equation with parameter

[mathwannabe]

Hello everybody

Since I have no way to check my answer, I am hoping my favorite forum will jump in and help me out

1) Set of all values of the real parameter \(\displaystyle m\) for which the roots of the quadratic equation \(\displaystyle (m-2)x^2-2mx+2m+2=0\) are real and with different signs is?

So, here is how I did it:

In order for the equation to have two solutions and for them to be real, \(\displaystyle D>0\).

\(\displaystyle D=4m^2-8(m-2)=4m^2-8m-16=m^2-2m-4\)

\(\displaystyle m^2-2-4>0\)

\(\displaystyle m<1-\sqrt{5}\) ^ \(\displaystyle m>1+\sqrt{5}\)

Also, for the the roots to have different signs, it must be that \(\displaystyle x1(x2)<0\)

\(\displaystyle x1(x2)=\dfrac{2m+2}{m-2}\)

\(\displaystyle \dfrac{2m+2}{m-2}<0\)

\(\displaystyle -1<m<2\)

My answer is empty set. There is no such m.

 

[mathwannabe]

hello everybody :d

since i have no way to check my answer, i am hoping my favorite forum will jump in and help me out

1) set of all values of the real parameter \(\displaystyle m\) for which the roots of the quadratic equation \(\displaystyle (m-2)x^2-2mx+2m+2=0\) are real and with different signs is?

So, here is how i did it:

In order for the equation to have two solutions and for them to be real, \(\displaystyle d>0\).

\(\displaystyle d=4m^2-8(m-2)=4m^2-8m-16=m^2-2m-4\) <<<<<<<<<< mistake

\(\displaystyle m^2-2-4>0\)

\(\displaystyle m<1-\sqrt{5}\) ^ \(\displaystyle m>1+\sqrt{5}\)

also, for the the roots to have different signs, it must be that \(\displaystyle x1(x2)<0\)

\(\displaystyle x1(x2)=\dfrac{2m+2}{m-2}\)

\(\displaystyle \dfrac{2m+2}{m-2}<0\)

\(\displaystyle -1<m<2\)

my answer is empty set. There is no such m.

i saw the mistake... Nevermind the answer

 

[mathwannabe]

Hello everybody

Since I have no way to check my answer, I am hoping my favorite forum will jump in and help me out

>>>>FIXED<<<<

1) Set of all values of the real parameter \(\displaystyle m\) for which the roots of the quadratic equation \(\displaystyle (m-2)x^2-2mx+2m+2=0\) are real and with different signs is?

So, here is how I did it:

In order for the equation to have two solutions and for them to be real, \(\displaystyle D>0\).

\(\displaystyle D=4m^2-4(m-2)(2m+2)=-4m^2+8m+16\)

\(\displaystyle -m^2+2m+4>0\)

\(\displaystyle 1-\sqrt{5}<m<1+\sqrt{5}\)

Also, for the the roots to have different signs, it must be that \(\displaystyle x1(x2)<0\)

\(\displaystyle x1(x2)=\dfrac{2m+2}{m-2}\)

\(\displaystyle \dfrac{2m+2}{m-2}<0\)

\(\displaystyle -1<m<2\)

My answer is : \(\displaystyle -1<m<2\).

This is a fixed answer

 

[mathwannabe]

Good! Nikola Tesla will be proud of you

Ha ha ha ha