SOLVING CUBIC EQUATIONS WITH ADDITIONAL INFORMATION OF ROOTS

Question 1 :

Solve the cubic equation : 2x³ − x² −18x + 9 = 0, if sum of two of its roots vanishes

Solution :

Sum of two roots vanishes means, sum of zeroes will be equal to 0.

Let the roots of the cubic equation will be α, - α and β.

From the given equation, we have 

a = 2, b = -1, c = -18 and d = 9

Sum of roots  =  -b/a

Sum of roots (α + (- α) + β) =  1/2

β =  1/2

Product of roots  =  -d/a  

Product of roots α (- α) β  =  -9/2

-α² β  =  -9/2

By applying the value of β, we get 

α² (1/2)  =  9/2

α²  =  9, α  =  ± 3

Hence the roots are 3, -3 and 1/2.

Question 2 :

Solve the equation 9x³ − 36x² + 44x −16 = 0 if the roots form an arithmetic progression.

Solution :

Let a - d, a and a + d be the roots of the equation which are in arithmetic progression.

Sum of roots  =  -b/a  =  36/9  =  4

a - d + a + a + d  =  4

3a  =  4  ==>  a  =  4/3

Product of roots  =  -d/a  =  16/9

(a - d) a (a + d)  =  16/9

a(a²  - d²)  =  16/9

By applying the value of a, we get 

(4/3) ((4/3)²  - d²)  =  16/9

((16/9) - d²)  =  (16/9) (3/4)

((16/9) - d²)  =  4/3

(16/9) - (4/3)  =  d²

d²  =  4/9

d  =  ± (2/3)

Hence the roots are 2/3, 4/3 and 2.

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