SOLVING CUBIC EQUATIONS WORD PROBLEMS

Problem 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 - 2 (c/a)

=  (b² - 2ca)/a²

So, the answer is (b² - 2ca)/a².

Problem 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

Problem 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

√(p/q) + √(q/p) + √(n/l) : 

=  √p/√q + √q/√p + √(n/l) 

=  (p + q)/√(pq)  + √(n/l)

  =  (-n/l) / √(n/l)  + √(n/l)

  =  - √(n/l)  + √(n/l)

  =  0

Problem 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'α² + p'q) - (p α² + q'p)  =  0

p'α² + p'q - p α² - q'p  =  0

α²(p'- p) + (p'q - q'p)  =  0

α²  =   (p'q - q'p) / (p'- p) 

α  =   (p'q - q'p) / (p'- p) α

α  =   (p'q - q'p) / (q - q')

Hence proved.

Problem 5 :

Formulate 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  =  x^1/3

x^1/3 + x  =  6

x^1/3  =  (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

 x³ - 18x² - 109x - 216  =  0

Problem 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  =  x^1/3

x^1/3 + x  =  12

x^1/3  =  (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

x³ - 36x² + 432x + x - 1728  =  0

x³ - 36x² + 433x - 1728  =  0

x³ + x  =  12

x³ + x - 12  =  0

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.