Find maximum integral value of k

[anemone]

Find the maximum integral value of $k$ such that

$(k-4)(x^2+x+1)^2-(k-3)(x^2+x+1)(x^2+x+2)+(x^2+x+2)^2=0$

has at least one real root for $x$.

 

[Greg]

Spoiler

\(\displaystyle (k-4)(x^2+x+1)^2-(k-3)(x^2+x+1)(x^2+x+2)+(x^2+x+2)^2=0\)

\(\displaystyle k(x^2+x+1)-4(x^2+x+1)^2-k(x^2+x+1)(x^2+x+2)+3(x^2+x+1)(x^2+x+2)+(x^2+x+2)^2=0\)

\(\displaystyle -k(x^2+x+1)+(x^2+x+1)[3x^2+3x+6-4x^2-4x-4]+(x^2+x+2)^2=0\)

\(\displaystyle -k(x^2+x+1)+(x^2+x+1)(2-x-x^2)+(x^2+x+2)^2\)

Expanding, cancelling like terms and rearranging gives

\(\displaystyle (5-k)x^2+(5-k)x+(6-k)=0\)

The above quadratic has roots for \(\displaystyle k\in\left(5,\frac{19}{3}\right]\), hence the maximum integral value of \(\displaystyle k\) for which the given equation has at least one root is \(\displaystyle 6\).

 

[anemone]

Thanks for participating, greg1313! Your answer is correct!