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.