# CUBIC EQUATION PRACTICE PROBLEMS

## About "Cubic Equation Practice Problems"

Cubic Equation Practice Problems :

Here we are going to see some practice problems on solving cubic equations.

## Cubic Equation Practice Problems - Practice questions

Question 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)

 x - 2  =  0x  =  2 x - 3  =  0x  =  3 3x - 1  =  03x  =  1x  =  1/3

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

General form of  polynomial of 4th degree.

ax4 + bx3 + cx2 + dx + e = 0

α + β + γ + δ  =  -b/a

α β +  α γ αδ + βγ + βδ + γ δ  =  c/a

α β γ + α β δ + α γ δ + βγδ   =  -d/a

α β γ δ  =  e/a

Question 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+d+ 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 + δ =  (α+β+γ+δ)2 - 2 (αβ+αγ+αδ+βγ+βδ+γδ

α2+ β2 + γ2 + δ =  (4)2 - 2 (3)

=  16 - 6

=  10

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

Question 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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