Devilish fun

Algebra Level 4

The product of the real roots of the polynomial

$f(x) = x^4 - 4x^3 + 6x^2 -4x-666$

can be written as $a - \sqrt{b}$ , where $a$ and $b$ are positive integers. What is the value of $a+b$ ?

The answer is 668.

2 solutions

Matthew Fan
Sep 24, 2013

We begin by noticing that this is similar to $(x-1)^4$ . So we have $f(x)=(x-1)^4-667$ So $(x-1)^4=667$ .

Since we want only the product of real roots, we have $(x-1)^2=\sqrt{667}$ by square-rooting on both sides. $x^2-2x+1-\sqrt{667}=0$ Product of roots= $1-\sqrt{667}$ by Vieta's Theorem. Hence $\boxed{668}$ is our final answer

Moderator note:

Nicely done!

Well I noticed that the coefficients of the function are :

1 (-4) 6 (-4) (and the constant)

Sound familiar? Look at Pascal's triangle and you'll see (1 4 6 4 1), which is awfully close to this, so I added 667 to both sides to make the function

$667 = x^4 - 4x^3 + 6x^2 - 4x + 1$

which then factors quite nicely to : $667 = (x-1)^4$

And then we can take the 4th root of each side to get :

$\pm \: \sqrt [4] {667} = (x-1)$

And solving gets : $x = 1 \pm \sqrt [4] {667}$

So the two real roots are $( 1 + \sqrt [4] {667})$ and $(1 - \sqrt[4] {667)}$

Multiplying both roots gets : $1 - \sqrt {667}$

And therefore the answer is : $\boxed{668}$

Omar Pulido - 7 years, 8 months ago

nicely done

Sampath Rachumallu - 7 years, 8 months ago

nicely solved!!! well done

Priyanshi Somani - 7 years, 8 months ago

well done I really appreciate this solution

Fatema Alzhraa Nasser - 7 years, 8 months ago

here in the question since they have mentioned that the product of real roots of x are in the form $a-\sqrt{b}$ we simplify and then apply Vieta's theorem. But if they would have not mentioned it we would think it $-666$ . So my question is why is it now $-666$ . Please clarify is it because the roots repeat.

shreyas S K - 7 years, 8 months ago

TOO AWESOME SOLUTION

Vaibhav Prasad - 6 years, 2 months ago

Nicky Sun
Sep 28, 2013

Using the fact that $(x-1)^4=x^4-4x^3+6x^2-4x+1$ , we have that $f(x)=(x-1)^4 - 667$ . Thus, the real roots of $f(x)$ are $1+\sqrt[4]{667}$ and $1-\sqrt[4]{667}$ . Thus, the product is $1-\sqrt{667}$ , and so $a+b =1+667=\boxed{668}$

Devilish fun

The answer is 668.

2 solutions

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