MHB Find maximum integral value of k

anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
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 $x$.
 
Mathematics news on Phys.org
$$(k-4)(x^2+x+1)^2-(k-3)(x^2+x+1)(x^2+x+2)+(x^2+x+2)^2=0$$

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

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

$$-k(x^2+x+1)+(x^2+x+1)(2-x-x^2)+(x^2+x+2)^2$$

Expanding, cancelling like terms and rearranging gives

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

The above quadratic has roots for $$k\in\left(5,\frac{19}{3}\right]$$, hence the maximum integral value of $$k$$ for which the given equation has at least one root is $$6$$.
 
Thanks for participating, greg1313! Your answer is correct!:cool:
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top