Proving No Intersection: y=2x-1 & y=x^4+3x^2+2x

[msimard8]

Heres the quesion

Prove that the line whose equation is y=2x-1 does not intersect the curve with equation y=x^4 + 3x^2 +2x.

We are suppose to solve this using indirect proof, thus assuming the equations do intersect, and proving that wrong.

i let the y's equal each other, but that isn't getting me anywhere

where should i start.

thanks

 

[radou]

Heres the quesion

Prove that the line whose equation is y=2x-1 does not intersect the curve with equation y=x^4 + 3x^2 +2x.

We are suppose to solve this using indirect proof, thus assuming the equations do intersect, and proving that wrong.

i let the y's equal each other, but that isn't getting me anywhere

where should i start.

thanks

Of course that is getting you somewhere. Ever thought of substituting x^2 = t ?

 

[Data]

Let $y_1(x) = 2x-1$ and $y_2(x) = x^4 + 3x^2 + 2x$. Look at $f(x) = y_2(x) - y_1(x) = x^4 + 3x^2 + 1$.

If $y_1$ and $y_2$ intersect at $x_0 \in \mathbb{R}$ then $f(x_0) = 0$. Can you get the contradiction (what do you know about $x^2, \, x^4$ when $x \in \mathbb{R}$?)?

 

[msimard8]

Of course that is getting you somewhere. Ever thought of substituting x^2 = t ?

umm yea i got the roots, x=1 or x=1 or x=-1

but what does that mean

 

[radou]

umm yea i got the roots, x=1 or x=1 or x=-1

but what does that mean

How did you get these roots? Let's start again. Intersection means setting x^4+3x^2+2x = 2x - 1, which implies x^4+3x^2+1=0. Now, as said, substitute x^2 = t, and solve the quadratic equation. Both solutions of this equation t1 and t2 are negative. So, substituting back to x^2 = t means that there is no real solution for the equation x^4+3x^2+1=0, i.e. y1 = x^4+3x^2+2x and y2 = 2x - 1 don't intersect.

 

[msimard8]

thank you so much


i just made a simple sign error in factoring t^2 + t +1

which gave me wrong roots

thanks again