Question

Sum of positive roots of the equation (x2 - 12x + 35) (x2 + 10x + 24) = 504 is

Answer 1

Answer:

(3)11

Step-by-step explanation:

We are given that

$(x^2-12x+35)(x^2+10x+24)=504$

We have to find the sum of positive roots of the equation.

$x^2(x^2+10x+24)-12x(x^2+10x+24)+35(x^2+10x+24)=504$

$x^4+10x^3+24x^2-12x^3-120x^2-288x+35x^2+350x+840=504$

$x^4-2x^3-61x^2+62x+840-504=0$

$x^4-2x^3-61x^2+62x+336=0$

Factor of 336

2,3,4,6,8,7,

Let x=2

$2^4-2^4-61(2^2)+62(2)+336\neq0$

x=2 is not the root of equation

x=-2

$(-2)^4-2(-2)^3-61(-2)^2+62(-2)+336=0$

Hence x=-2 is the root of equation.

x+2 is a factor of equation.

x=3

$3^4-2(3^3)-61(3^2)+62(3)+336=0$

Therefore, x=3  is the root of equation.

$(x+2)(x-3)(x^2-x-56)=0$

$(x+2)(x-3)(x^2-8x+7x-56)=0$

$(x+2)(x-3)(x(x-8)+7(x-8))=0$

$(x+2)(x-3)(x-8)(x+7)=0$

$x-8=0\implies x=8$

$x+7=0\implies x=-7$

Positive roots are 3 and 8

Sum of positive roots=3+8=11

Option (3) is true.