CUBIC EQUATION PRACTICE PROBLEMS

Problem 1 :

Solve the equation 3x³ −16x² + 23x − 6 = 0 if the product of two roots is 1.

Solution :

Let us solve the given cubic equation using synthetic division.

We ge the zero remainder by applying the value of x as 2. So, (x - 2) is a factor.

The factors are (x - 2) (3x² - 10x + 3)

By factoring the quadratic equation, we get

  =  3x² - 1x - 9x + 3

  =  x (3x - 1) - 3(3x - 1)

  =  (x - 3) (3x - 1)

Hence the values of x are 2, 3 and 1/3.

Problem 2 :

Find the sum of squares of roots of the equation :

2x⁴ - 8x³ + 6x² - 3  =  0

Solution :

By comparing the given equation with general form of polynomial of degree 4, we get 

a = 2, b = -8, c = 6 and d = -3

α+β+γ+δ  =  -b/a  =  8/2  =  4

αβ+αγ+αδ+βγ+βδ+γδ  =  c/a  =  6/2  =  3

We have to find 

Applying the algebraic identity α² + β² + γ² + δ² 

(a+b+c+d)² = a²+b²+c²+d² + 2(ab + ac + ad + bc + bd + cd)

we get

(α+β+γ+δ)² = α²+ β² + γ² + δ² + 2 (αβ+αγ+αδ+βγ+βδ+γδ) 

α²+ β² + γ² + δ²  =  (α+β+γ+δ)² - 2 (αβ+αγ+αδ+βγ+βδ+γδ) 

α²+ β² + γ² + δ²  =  (4)² - 2 (3) 

  =  16 - 6

  =  10

Hence the sum of squares of roots of the equation is 10.

Problem 3 :

Solve the equation x³ - 9x² + 14x + 24 = 0 if it is given that two of its roots are in the ratio 3: 2.

Solution :

-1 is one of the roots of the cubic equation.By factoring the quadratic equation x² - 10x + 24, we may get the other roots.

x² - 10x + 24  =  x² - 6x - 4x + 24

  =  x(x - 6) - 4(x - 6)

  =  (x - 4) (x - 6)

x - 4  =  0 and x - 6  =  0

x  =  4 and x = 6

Hence the roots of the cubic equation are -1, 4 and 6.

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