Answer:
(3)11
Step-by-step explanation:
We are given that
[tex](x^2-12x+35)(x^2+10x+24)=504[/tex]
We have to find the sum of positive roots of the equation.
[tex]x^2(x^2+10x+24)-12x(x^2+10x+24)+35(x^2+10x+24)=504[/tex]
[tex]x^4+10x^3+24x^2-12x^3-120x^2-288x+35x^2+350x+840=504[/tex]
[tex]x^4-2x^3-61x^2+62x+840-504=0[/tex]
[tex]x^4-2x^3-61x^2+62x+336=0[/tex]
Factor of 336
2,3,4,6,8,7,
Let x=2
[tex]2^4-2^4-61(2^2)+62(2)+336\neq0[/tex]
x=2 is not the root of equation
x=-2
[tex](-2)^4-2(-2)^3-61(-2)^2+62(-2)+336=0[/tex]
Hence x=-2 is the root of equation.
x+2 is a factor of equation.
x=3
[tex]3^4-2(3^3)-61(3^2)+62(3)+336=0[/tex]
Therefore, x=3 is the root of equation.
[tex](x+2)(x-3)(x^2-x-56)=0[/tex]
[tex](x+2)(x-3)(x^2-8x+7x-56)=0[/tex]
[tex](x+2)(x-3)(x(x-8)+7(x-8))=0[/tex]
[tex](x+2)(x-3)(x-8)(x+7)=0[/tex]
[tex]x-8=0\implies x=8[/tex]
[tex]x+7=0\implies x=-7[/tex]
Positive roots are 3 and 8
Sum of positive roots=3+8=11
Option (3) is true.
