MHB Finding 2d-2c from Equation with 4 Roots

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The equation (x-2)(x+1)(x+4)(x+7)-19=0 is transformed into a polynomial form, resulting in x^4+10x^3+15x^2-50x-75. The roots of this polynomial are calculated as x=1/2(-5±√5) and x=1/2(-5±√85). The values of a, b, c, and d are identified with a<b<c<d. The final expression for 2d-2c is derived as √85-√5, which simplifies to √5(√17-1). The solution is confirmed as correct by participants in the discussion.
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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)
 
Thread 'Erroneously finding discrepancy in transpose rule'

Thread 'Erroneously finding discrepancy in transpose rule'

Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...
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