Sum of roots #1

Algebra Level 4

$\large \left( x+1 \right) \left( x+2 \right) \left( x+3 \right) =\frac { 720 }{ \left( x+4 \right) \left( x+5 \right) \left( x+6 \right)}$

Find sum of all real solutions of the equation above.

1 solution

Kushal Bose
May 1, 2017

$(x+1)(x+2)(x+3)(x+4)(x+5)(x+6)=720$

Now start grouping $(x+1)(x+6) \times (x+2)(x+5) \times (x+3)(x+4)=720 \\ (x^2+7 x+6)((x^2+7 x+10))(x^2+7 x+12)=720$

Assume $x^2+7x=p$

Now equation becomes $(p+6)(p+10)(p+12)=720 \\ p^3 +28p^2+252p +720=720 \\ p(p^2+28p+252)=0$

So, one solution is $p=0 \implies x^2+7x=0 \implies x=0,-7$

Next quadratic part has negative discriminant so there exists no real solution.

Sum of real roots $0+(-7)=-7$

Exact same procedure! Nice solution :) Btw, you made a typo in the second line

$(x+1)(x+6) \times (x+2)(x+{\color{#D61F06}5}) \times (x+3)(x+4) = 720$

Tapas Mazumdar - 4 years, 1 month ago

Thanks, I hve fixd it

Kushal Bose - 4 years, 1 month ago