Inspired by some flawed solution

Algebra Level 4

Let $a,b,c$ and $d$ be integer roots to the equation $x^4 + ax^3 + bx^2 + cx + d = 0$ . Find the product $a\times b \times c \times d$ .

The answer is 0.

1 solution

Tay Yong Qiang
May 4, 2016

By Vieta's Formula,

$abcd=d$

$d(abc-1)=0$

$\therefore d=0$ or $abc=1$

Let's investigate the case where $abc=1$

By Vieta's Formula, $ab+bc+cd+ad=b$

$d=\frac{b-ab-bc}{a+c}\Rightarrow a+c\neq0\Rightarrow a\neq-c$

$\therefore a=c=b=1\Rightarrow d=\frac{1-1-1}{1+1}=-\frac{1}{2}\not\in\mathbb{Z}$

$a=c=-1, b=1\Rightarrow d=\frac{1-1(-1)-1(-1)}{-1+(-1)}=-\frac{3}{2}\not\in\mathbb{Z}$

$\therefore abc\neq1$

$\therefore d=0, a\times b\times c\times d=\boxed{0}$

You have only shown that this is true for $d=0$ . What about for $d\ne0$ ?

Pi Han Goh - 5 years, 1 month ago

If $d\neq 0$ , then $abc-1=0\Rightarrow abc=1$ , which is not possible as i have stated above. Perhaps my formatting is a bit off lol, the word 'or' is just floating in the middle of my solution, but it should mean $d=0$ OR $abc=1$ .

Tay Yong Qiang - 5 years, 1 month ago

Why is $abc = 1$ not possible?

Pi Han Goh - 5 years, 1 month ago

@Pi Han Goh – So sorry, I've overlooked the question.

By Vieta's Formula, $ab+bc+cd+ad=b$

$d=\frac{b-ab-bc}{a+c}\Rightarrow a+c\neq0\Rightarrow a\neq-c$

$\therefore a=c=b=1\Rightarrow d=\frac{1-1-1}{1+1}=-\frac{1}{2}\not\in\mathbb{Z}$

$a=c=-1, b=1\Rightarrow d=\frac{1+1(-1)-1(-1)}{-1+(-1)}=-\frac{1}{2}\not\in\mathbb{Z}$

$\therefore abc\neq1$

Tay Yong Qiang - 5 years, 1 month ago

@Tay Yong Qiang – Great work! =D

You should add that to your solution! +1

Pi Han Goh - 5 years, 1 month ago

@Pi Han Goh – Added, thanks for pointing out my careless mistake.

Tay Yong Qiang - 5 years, 1 month ago

@Tay Yong Qiang – Haha! no problem! we're all learning after all! ;)

Pi Han Goh - 5 years, 1 month ago

I have edited my answer due to bad formatting, my apologies im kinda new to LaTeX.

Tay Yong Qiang - 5 years, 1 month ago

This line is wrong." By Vieta's Formula $ab + bc + cd + ab = b$ ". It should be $ab + bc + cd + ad + ac + bd = b$ . The coefficient of $x^2$ is equal to the sum of pairwise products of roots.

Krutarth Patel - 6 months, 2 weeks ago

Inspired by some flawed solution

The answer is 0.

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