Almost Quartic Equation

It is given that:

$x^3+x^2+x+1 = 20^2$

$x^3+x^2+x-399 = 0$

$(x-7)(1+8x+57) = 0$

$(x-7)\left(\frac {-8+\sqrt{64-4(57)}}{2}\right) \left(\frac {-8-\sqrt{64-4(57)}}{2}\right) = 0$

$(x-7)(-4+i\sqrt{41})(-4-i\sqrt{41}) = 0$

$\Rightarrow a = 4$ , $b = 41$ and $a+b = \boxed{45}$

Well the solution is as follows

one of the root is x-7 so dividing the eqn by x-7 we get x^2+8x+57.On solving it we get the required root as

-4-√41 so a=4 and b=41 a+b=45

But it should not be a level 4 Parth Lohomi

multiply and divide by (x-1) then numerator becomes x^4-1 and denominator as x-1 now om simplifying we get [x^4-400x+399=0] whose real roots are 1 and 7.now rest is simple.

Rewrite the equation as $x^3+3x^2-2x^2+3x-2x+1=400$

With the help of Pascals triangle we can write it as $(x+1)^3-2x(x+1)=(x+1)((x+1)^2-2x)=(x+1)(x^2+1)=400$

This tells us that the factors of $400$ can't differ too much. This rules out some factors.

By a few trials, we find that $x=7$ ,which is one of the solutions.

Using Vieta's formulas for the original equation ( $x^3+x^2+x-399=0$ ):

$7ab=399$

$7+a+b=-1$

( $7,a,b$ are the roots)

(Those who don't know the Vieta's formulas, search it online.)

Solve these equations to get the other 2 roots, which are $-4\pm i\sqrt{41}$