Quadratic equation conditions

[Loki123]

Where did I go wrong???

 

[Loki123]

It would not let me post the final pic

 

[blamocur]

It looks to me that you teacher is wrong: at [imath]\alpha = \frac{\pi}{3}[/imath] we only get [imath]x=-\frac{1}{2}[/imath], which is not at all positive. In addition, your teacher cannot think outside the [imath][0, 2\pi][/imath] box

 

[Loki123]

It looks to me that you teacher is wrong: at [imath]\alpha = \frac{\pi}{3}[/imath] we only get [imath]x=-\frac{1}{2}[/imath], which is not at all positive. In addition, your teacher cannot think outside the [imath][0, 2\pi][/imath] box

But when we plug - 1/2 where cos a is we get - 2+sqrt(4+4)=-2+2sqrt(2), which I think is positive

 

[blamocur]

Not sure why you brought up [imath]\cos\alpha= - \frac{1}{2}[/imath], for which the two roots have different signs.
My point is that you teacher is wrong about two positive roots. Quadratic equation [imath]2x^2+2x+c=0[/imath] cannot have two positive roots for any value of [imath]c[/imath] -- don't you agree?

 

[Loki123]

Not sure why you brought up [imath]\cos\alpha= - \frac{1}{2}[/imath], for which the two roots have different signs.
My point is that you teacher is wrong about two positive roots. Quadratic equation [imath]2x^2+2x+c=0[/imath] cannot have two positive roots for any value of [imath]c[/imath] -- don't you agree?

I do.