Elementary Algebra

Algebra Level 1

Two of the roots of the equation 2x^3-3x^2+px+q = 0 are 3 and -2. Find the third root of the equation.

Enter answer in decimal.


The answer is 0.5.

Aryan Gaikwad by Aryan Gaikwad , 22, India

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2 solutions

Kajal Choudhary
Feb 22, 2014

Given equation 2x^3-3x^2+px+q = 0 has three roots ,two of them are 3 and -2. Let's assume 3rd root is a. For any cubic equation ax^3 + bx^2 + c^x + d ,three roots are p,q,r then p+q+r = - (b/a) ; p q+q r+p r = c/a ; p q*r = -(d/a) So by using these results a+3-2= 3/2 =>a=1/2

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Aryan Gaikwad
Feb 15, 2014

2x^3-3x^2+px+q = 0 ; substitute 3 for x ; 54-27+3p+q = 0 ; substitute -2 for x ; -16-12-2p+q = 0 ;

27+3p+q = 0 ; -28-2p+q = 0 ; 55+5p = 0 ; p=-11 ; q=6 ;

2x^3-3x^2-11x+6 = 0 ; (x-3) (x+2) (2 x-1) = 0 ;

x = 1/2

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we know that the sum of roots in ax^3+bx^2+px+q = 0 is -b/a;
hence here 3-2+ y ( third root) = 3/2; y= 1.5 - 1 = .5

Rahul Mishra - 7 years, 3 months ago

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