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#### Albert

##### Well-known member

- Jan 25, 2013

- 1,225

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=0

and a<b<c<d

find : 2d-2c

- Thread starter Albert
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- Thread starter
- #1

- Jan 25, 2013

- 1,225

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=0

and a<b<c<d

find : 2d-2c

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- #2

- Feb 7, 2012

- 2,792

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

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

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- #3

- Jan 25, 2013

- 1,225

Opalg :good solution