Roots finding 24x^3 -52x^2 -60x = 0

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should factor out x such that:

x(24x^2 -52x -60) = 0

You need to solve two equations x = 0 and 24x^2 -52x -60 = 0 such that:

24x^2 -52x -60 = 0

12x^2 - 26x - 30 = 0

6x^2 - 13x - 15 = 0

You need to complete the square such that:

6x^2 - 13x + 169/24 - 169/24 - 15 = 0

(sqrt6*x - 13/(2sqrt6))^2 = 169/24 + 15

(sqrt6*x - 13/(2sqrt6))^2 = 529/24

(sqrt6*x - 13/(2sqrt6)) = +- 23/(2sqrt6)

sqrt6*x - 13/(2sqrt6)) = 23/(2sqrt6) => x = 36/12 = 3

sqrt6*x - 13/(2sqrt6)) = -23/(2sqrt6) => x = -10/12 = -5/6

Hence, evaluating the solutions to the given equation yields x=-5/6; x=0 ; x = 3.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

The equation will have 3 roots, since the maximum order of polynomial is 3.

We'll factorize by x and we'll get:

x(24x^2 - 52x - 60) = 0

We'll set each factor as zero:

x1 = 0

and

24x^2 - 52x - 60 = 0

We'll divide by 4:

6x^2 - 13x - 15 = 0

We'll apply the quadratic formula:

x1 = [-b+sqrt(b^2 - 4ac)]/2a

x1 = [13+sqrt(169+360)]/12

x1 = (13+23)/12

x1 = 3

x2 = (13-23)/12

x2 = -10/12

x2 = -5/6

The roots of the equation are: {-5/6 , 0 , 3}.

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