Question Number 11501 by @ANTARES_VY last updated on 27/Mar/17

(4x^2 −7x−5)(5x^2 +13x+3)(3x−x^2 −8)=0.  find  all  the  multiples  of  the  real  roots  of  the  equation.

Commented bymrW1 last updated on 27/Mar/17

do you mean the product of all real  roots of the equation?    then the answer is ((−5×3)/(4×5))=−(3/4).

Answered by Joel576 last updated on 27/Mar/17

• 4x^2  − 7x − 5 = 0      x_(1,2)  = ((7 ± (√(49 − (4.4.−5))))/(2.4))              = ((7 ± (√(129)))/8)      x_1  = ((7 + (√(129)))/8)      x_2  = ((7 − (√(129)))/8)  • 5x^2  + 13x + 3 = 0      x_(3,4)  = ((−13 ± (√(169 − (4.5.3))))/(2.5))              = ((−13 ± (√(109)))/(10))       x_3  = ((−13 + (√(109)))/(10))       x_4  = ((−13 − (√(109)))/(10))  • −x^2  + 3x − 8 = 0      x_(5,6)  = ((−3 ± (√(9 − (4.−1.−8))))/(2.−1))              = ((−3 ± (√(−23)))/(−2))  → imaginary     (((7 + (√(129)))/8) . ((7 − (√(129)))/8)) (((−13 + (√(109)))/(10)) . ((−13 − (√(109)))/(10)))  = (((49 − 129)/(64)))(((169 − 109)/(100)))  = −(3/4)

Commented by@ANTARES_VY last updated on 27/Mar/17

error

Answered by sandy_suhendra last updated on 27/Mar/17

I′ve made a mistake, so I corrected  4x^2 −7x−5=0 has 2 real roots x_1  and x_(2 ) because D>0  5x^2 +13x+3=0 has 2 real roots x_3  and x_4  because D>0        3x−x^2 −8=0 has 2 imaginary roots because D<0  so (4x^2 −7x−5)(5x^2 +13x+3)=0 has 4 real roots       coefisien of x^4 =a  coefisien of x^3 =b  coefisien of x^2 =c  coefisien of x^ =d  constant = e  so e=−5×3=−15       a = 4×5=20  x_1 .x_2 .x_3 .x_4 =(e/a)=((−15)/(20))=− (3/4)

Answered by ajfour last updated on 27/Mar/17

(3/4)n  where n belongs to integers.