Finding 2d-2c from Equation with 4 Roots

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The discussion focuses on finding the expression 2d-2c from the roots of the polynomial equation (x-2)(x+1)(x+4)(x+7)-19=0. The equation simplifies to x^4+10x^3+15x^2-50x-75, which factors into (x^2+5x+5)(x^2+5x-15). The roots are determined to be x=1/2(-5±√5) and x=1/2(-5±√85). Consequently, the value of 2d-2c is calculated as √85-√5, which can be expressed as √5(√17-1).

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a,b,c,d are four roots of equation :

(x-2)(x+1)(x+4)(x+7)-19=0

and a<b<c<d

find : 2d-2c
 
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Re: find :2d-2c

Albert said:
a,b,c,d are four roots of equation :

(x-2)(x+1)(x+4)(x+7)-19=0

and a<b<c<d

find : 2d-2c
$(x-2)(x+1)(x+4)(x+7)-19 = x^4+10x^3+15x^2-50x-75 = (x^2+5x+5)(x^2+5x-15).$

The roots are $x=\frac12(-5\pm\sqrt5)$ and $x=\frac12(-5\pm\sqrt{85})$, so that $2d-2c = \sqrt{85}-\sqrt5 = \sqrt5(\sqrt{17} - 1).$
 
Re: Find :2d-2c

Opalg :good solution(Yes)
 

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