# SOLVING CUBIC EQUATIONS WORD PROBLEMS

## About "Solving Cubic Equations Word Problems"

Solving Cubic Equations Word Problems :

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

## Solving Cubic Equations Word Problems - Practice questions

Question 1 :

If α, β  and γ are the roots of the polynomial equation ax3 + bx2 + cx + d = 0 , find the Value of ∑ α/βγ in terms of the coefficients

Solution :

∑ α/βγ  =  (α/βγ) + (β/γα) + (γ/αβ)

=  (α2+ β2 + γ2)/αβγ

=  (α + β + γ)2 2 (αβ+ βγ αγ)  ----(1)

α + β + γ  =  -b/a

αβ+ βγ + αγ  =  c/a

By applying the above values in (1), we get

=  (-b/a)2 (c/a)

=  (b2 - 2ca)/a2

Hence the answer is (b2 - 2ca)/a2.

Question 2 :

If α, β, γ and δ are the roots of the polynomial equation 2x4 + 5x3 − 7x2 + 8 = 0 , find a quadratic equation with integer coefficients whose roots are α + β + γ + δ and αβγδ.

Solution :

2x4 + 5x3 − 7x2 + 8 = 0

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

a  =  2, b = 5, c = -7, d = 0 and e = 8.

α + β + γ + δ  =  -b/a  =  -5/2

αβγδ  =  e/a  =  8/2  =  4

Let us look into the next problem on "Solving Cubic Equations Word Problems".

Question 3 :

If p and q are the roots of the equation lx2 + nx + n = 0, show that √(p/q) + √(q/p) + √(n/l)  =  0

Solution :

Since p and q are the roots of the equation, let us find sum of roots and product of roots of the given quadratic equation.

Sum of roots  =  p + q  =  -n/l

Product of roots pq  =  n/l

L.H.S

√(p/q) + √(q/p) + √(n/l)

=  (p + q)/√(pq)  + √(n/l)

=  (-n/l) / √(n/l)  + √(n/l)

=  √(n/l)  + √(n/l)

=  0

R.H.S

Hence proved.

Let us look into the next problem on "Solving Cubic Equations Word Problems".

Question 4 :

If the equations x2 + px + q = 0 and x2 + p'x + q' = 0 have a common root, show that it must be equal to (pq'-p'q)/q q' or (q - q') / (p' - p)

Solution :

Let "α" be the common root

By applying "α" instead of "x", we get

α2 + pα + q = 0 -----(1)

α2 + p'α + q' = 0 ------(2)

(1) - (2)

(pα + q) - (p'α + q')  =  0

pα + q - p'α - q'  =  0

(p - p')α + (q - q')  =  0

α  =  - (q - q')/(p - p')

α  =   (q - q')/(p' - p)

To get the value of other root,

(1) x p' ==> p'α2 + pp'α + p'q = 0

(2) x p ==> p α2 + p'p α + q'p = 0

By subtracting these two equations, we get

(p'α+ p'q) - (α2 + q'p)  =  0

p'α+ p'q - α2 - q'p  =  0

α2(p'p) + (p'q - q'p)  =  0

α=   (p'q - q'p) / (p'p)

α  =   (p'q - q'p) / (p'- p) α

α  =   (p'q - q'p) / (q - q')

Hence proved.

Let us look into the next problem on "Solving Cubic Equations Word Problems".

Question 5 :

Formalate into a mathematical problem to find a number such that when its cube root is added to it, the result is 6.

Solution :

Let "x" be a required number.

Its cube root  =  x1/3

x1/3 + x  =  6

x1/3  =  (6 - x)

Taking cubes on both sides, we get

x  =  (6 - x)3

By applying the algebraic identity for (a - b)3, we get

(a - b)3  =  a3 - 3a2 b + 3ab2 - b3

x  =  63 - 3(36)x + 3(6)x2 - x3

x  =  216 - 108x + 18x2 - x3

x3 18x2 - 108x - x - 216  =  0

x18x2 - 109x - 216  =  0

Question 6 :

A 12 metre tall tree was broken into two parts. It was found that the height of the part which was left standing was the cube root of the length of the part that was cut away. Formulate this into a mathematical problem to find the height of the part which was cut away

Solution :

Let "x" be the broken part, its cube root  =  x1/3

x1/3 + x  =  12

x1/3  =  (12 - x)

Taking cubes on both sides, we get

x  =  (12 - x)3

By applying the algebraic identity for (a - b)3, we get

(a - b)3  =  a3 - 3a2 b + 3ab2 - b3

x  =  123 - 3(144)x + 3(12)x2 - x3

x  =  1728 - 432x + 36x2 - x3

x- 36x2 + 432x + x - 1728  =  0

x- 36x2 + 433x - 1728  =  0

x+ x  =  12

x+ x - 12  =  0

After having gone through the stuff given above, we hope that the students would have understood, "Solving Cubic Equations Word Problems".

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