Question

Solution please?

Answer 1

`5 <= a <= 19/3`

Explanation:

Given:

`(x^2+x+2)^2-(a-3)(x^2+x+2)(x^2+x+1)+(a-4)(x^2+x+1)^2 = 0`

Writing `t` for `x^2+x+1`, this becomes:

`0 = (t+1)^2-(a-3)(t+1)t+(a-4)t^2`

`color(white)(0) = (t^2+2t+1)-(a-3)(t^2+t)+(a-4)t^2`

`color(white)(0) = color(blue)(cancel(color(black)(t^2)))+2t+1-color(red)(cancel(color(black)(at^2)))-at+color(blue)(cancel(color(black)(3t^2)))+3t+color(red)(cancel(color(black)(at^2)))-color(blue)(cancel(color(black)(4t^2)))`

`color(white)(0) = (5-a)t+1`

`color(white)(0) = (5-a)(x^2+x+1)+1`

`color(white)(0) = (5-a)x^2+(5-a)x+(6-a)`

This has discriminant:

`Delta = (5-a)^2-4(5-a)(6-a)`

`color(white)(Delta) = (5-a)((5-a)-4(6-a))`

`color(white)(Delta) = (5-a)(3a-19)`

Hence `Delta >= 0` if and only if `5 <= a <= 19/3`