Sum It Up!

Algebra Level 2

The roots of the equation ${ 3x }^{ 2 }({ x }^{ 2 }+8)+16({ x }^{ 3 }-1)=0$ are not all unequal : The sum of all 4 roots is $-\frac { a }{ b }$ .

where a and b are Co Primes. Find a + b

2 solutions

Rindell Mabunga
Mar 4, 2014

$3x^2(x^2 + 8) + 16(x^3 -1$ = 3x^4 + 16x^3 + 24x^2 -16 = 0)

Therefore by Vieta's Theorem, the sum of all 4 roots will be

$-16/3$

$a = 16$ and $b = 3$

Therefore,

$a + b = 16 + 3 = 19$

Aldrin Vic Payuyo
Mar 8, 2014

roots are 4, -4, -16/6 & -16/6 adding the roots is equal to -16/3 a=16 ; b=3 therefore, a + b = 19