**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.

**Question 1 :**

Solve the equation 3x^{3} − 26x^{2} + 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 + r^{2})/r] = 26/3 ---(1)

(a/r) a (ar) = 8

a^{3} = 8, a = 2

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

2[(1 + r + r^{2})/r] = 26/3

[(1 + r + r^{2})/r] = 13/3

3r^{2} + 3r + 3 = 13r

3r^{2} - 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 2x^{3} − 6x^{2} + 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 x^{2}/coefficient of x^{3}

= 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.

2x^{2} - 2x - 1

x = (-b ± √b^{2} - 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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