Problem:
Find all real numbers $a$ for which the equation, $x^3 + ax^2 + 51x + 2023=0$, has two equal roots.
- This problem is from an algebra round of a local high school math competition that has already ended
My Work:
To find when $x^3 + ax^2 + 51x + 2023$ has two equal roots, we can use the properties of the discriminant. For a cubic polynomial, the discriminant is given by the following formula:
$Δ = 18abcd – 4b^3d + b^2c^2 – 4ac^3 – 27a^2d^2$
In our case, the cubic polynomial is $x^3 + ax^2 + 51x + 2023$, so $a = 1$, $b = a$, $c = 51$, and $d = 2023$.
If the polynomial has two equal roots, its discriminant must be equal to zero. So, we can set Δ equal to zero and solve for a:
$0 = 18(1)(a)(51)(2023) – 4a^3(2023) + a^2(51)^2 – 4(1)(51)^3 – 27(1)^2(2023)^2$
However, after this step, I am really stuck. I would appreciate some help! Thank you!
Best Answer
If it has two equal roots, it means all three roots are real, let the double roots at $x=c$, so
$$x^3 + ax^2 + 51x + 2023=(x-c)^2(x-b)=x^3-(2c+b)x^2+(c^2+2bc)x-bc^2$$
Compare coefficients
$$-bc^2=2023\tag{1}$$ $$c^2+2bc=51\tag{2}$$ $$b+2c=-a\tag{3}$$
From (1), (2), eliminate $b$, and we get
$$0=c^3 - 51 c - 4046=(-17 + c) (238 + 17 c + c^2)$$ Therefore, $c=17, b=-7$
From (3), we get $a=-27$