Question

The number of integers $k$ for which the equation $x^3-27x+k=0$ has at least two distinct integer roots is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

$x^3-27x+k=0$

Let a,b be the two distinct root of given equation $x^3-24x+k=0$

$\therefore a^3-27a+k=0$ ______ (1)

$b^3-27b+k=0$ _______ (2)

subtracting (1) and (2)

$a^3-b^3-27a+2+b=0$

$(a-b)(a^2+ab+b^2)-27(a-b)=0$

$\therefore (a-b)(a^2+ab+b^2-27)=0$

So $a-b=0;a=b$

OR $a^2+ab+b^2-27=0$

$\therefore 3b^2=27$

$\therefore b=\pm 3$

Put $b=+3$ in equation (2)

$3^3-27\left(3\right)+k=0$

$\therefore 27-81+k=0$

$\therefore k=8-27$

$\therefore k=54$

Put $b=-3$ in equation (2)

$(-3{)}^3-27(-3)+k=0$

$\therefore -27+81+k=0$

$\therefore k=-54$

$\therefore$ The number of integers are two.