SOLVING CUBIC EQUATIONS WHEN ADDITIONAL INFORMATION OF ROOTS ARE GIVEN

About "Solving Cubic Equations when Additional Information of Roots are Given"

Solving Cubic Equations when Additional Information of Roots are Given :

Here we are going to see some example problems of solving cubic with additional information of roots.

Solving Cubic Equations when Additional Information of Roots are Given - Practice questions

Question 1 :

Solve the equation 3x3 − 26x2 + 52x − 24 = 0 if its roots form a geometric progression.

Solution :

Let the roots of the cubic equation be a/r, a and ar.

By comparing the given equation with general form of cubic equation, we get 

a = 3, b = -26, c = 52 and d = -24.

In any cubic equation ,

Sum of roots  =  -b/a  =  26/3

Product of roots  =  -d/a  =  24/3  =  8

(a/r) + a + ar  =  26/3 

a[(1 + r + r2)/r]  =  26/3  ---(1)

(a/r) a (ar)  =  8

a3  =  8, a = 2

By applying the value of a in (1), we get

2[(1 + r + r2)/r]  =  26/3

[(1 + r + r2)/r]  =  13/3

3r2 + 3r + 3  =  13r

3r2 - 10r + 3  =  0

r = 1/3, r = 3

When a = 2 and r = 3

a/r = 2/3, a = 2 and ar  =  2(3) = 6

Hence the roots are 2/3, 2 and 6.

Question 2 :

Determine k and solve the equation 2x3 − 6x2 + 3x + k = 0 if one of its roots is twice the sum of the other two  roots.

Solution :

Let a, b and c be the roots of the cubic equation.

a = 2(b + c)

Sum of roots  =  -Coefficient of x2/coefficient of x3

  =  6/2  =  3

a + b + c  =  3

2b + 2c + b + c  =  3

3(b + c)  =  3

b + c  =  1

c = 1 - b

By applying the value of c in a = 2(b + c), we get

a  =  2(b + 1 - b)

a  =  2

k - 2  =  0

Hence the value of k is 2.

2x2 - 2x - 1

x = (-b ± √b2 - 4ac)/2a

x = (2 ± √4 + 8)/4

x = (2 ± 2√3)/4

x = (1 ± √3)/2

The required roots are 2, (1 ± √3)/2.

After having gone through the stuff given above, we hope that the students would have understood, "Solving Cubic Equations when Additional Information of Roots are Given".

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