Polynomial Q: roots of x^2+bx+c=0 differ by 2; prove b^2=4(1+c)

[Andy005]

The roots of the equation x²+bx+c=0 differ by 2. Prove that b²= 4(1+c).

I'm unsure on how to start this.

I made an attempt and I know it's incorrect but, I'm unsure if it's the correct approach.

View attachment 7971

 

[stapel]

The roots of the equation x²+bx+c=0 differ by 2. Prove that b²= 4(1+c).

I'm unsure on how to start this.

I made an attempt and I know it's incorrect but, I'm unsure if it's the correct approach.

View attachment 7971

I'm sorry, but your attachment is not displaying. Kindly please repost, starting from the beginning.

You plugged the given equation into the Quadratic Formula. You noted that the difference of the two roots will be created by the "plus/minus" on the square root. You used the given relation to create an equation. And... then what?

Please be complete. Thank you!

 

[HallsofIvy]

Take the smaller root to be p so that the larger is p+ 2. Then we have (x- p)(x- p- 2)= x^2- (2+ 2p)x+ p^2+ 2p= x^2+ bx+ c. Since those are to be true for all x, we must have 2+ 2p= -b and p^2+ 2p= c. Now calculate both b^2 and 4(1+ c) in terms of p.