# SOLVING POLYNOMIALS WITH RATIONAL POWERS

## About "Solving Polynomials with Rational Powers"

Solving Polynomials with Rational Powers :

Here we are going to see some example problems of solving polynomials.

## Solving Polynomials with Rational Powers - Practice questions

Question 1 :

Solve 8 x3/2n - 8 x-3/2n  =  63

Solution :

8 x3/2n - 8 x-3/2n  =  63

8 x3/2n - 8 (1/x3/2n)  =  63

x3/2n - (1/x3/2n)  =  (63/8)

Let t = x3/2n

t - (1/t)   =  63/8

8(t2 - 1)  =  63t

8t2 - 63t - 8  =  0

8t- 64t + t - 8  =  0

8t(t - 8) + 1(t - 8)  =  0

(8t + 1)(t - 8)  =  0

t  =  -1/8, t = 8

x3/2n  =  8

Raise 2/3 as power for both sides

(x3/2n)^2/3  =  (23)^2/3

x1/n  =  22

x1/n  =  4

x  =  4n

Question 2 :

Solve : Solution : Now let us solve the quadratic equation in terms of "t"

t = -b ± √(b2 - 4ac) / 2a

t  =  [(b2 + 6a2) ± √(b2 + 6a2)2 - 4(2ab)(3ab)]/2(2ab)

t  =  [(b2 + 6a2) ± √(b4 + 36a4 + 12a2b2 - 24a2b2)/4ab

t  =  [(b2 + 6a2) ± √(b2-6a2)2]/4ab

t  =  [(b2 + 6a2) ± (b2-6a2)]/4ab

t  =  [(b2 + 6a2) + (b2-6a2)]/4ab

 t = (b2+6a2)+(b2-6a2)/4abt  =  2b2/4abt  =  b/2a√(x/a)  =  tx/a  =  (b/2a)2x  =  ab2/4a2x  =  b2/4a t = (b2+6a2)-(b2-6a2)/4abt  =  12a2/4abt  =  3a/b√(x/a)  =  tx/a  =  (3a/b)2x  = 9a3/b2

Question 3 :

Solve the equations

(i) 6x4 − 35x3 + 62x2 − 35x + 6 = 0

Solution :

By suing synthetic division repeatedly, we may solve this problem. 2 and 3 are the solutions of the given polynomial. To know the other solutions, let us solve the quadratic equation.

6x2 - 5x + 1  =  0

6x2 - 2x - 3x + 1  =  0

2x (3x - 1) - 1(3x - 1)  =  0

(2x - 1) (3x - 1)  =  0

2x - 1  =  0 and 3x - 1  =  0

x  =  1/2  and x  =  1/3

Hence the solutions are 2, 3, 1/2 and 1/3

(ii)  x4 + 3x3 - 3x - 1 1 and -1 are the solutions of the given polynomial, to find other two solutions, let us solve the quadratic equation

x2 + 3x + 1  =  0

x = -b ± √(b2 - 4ac) / 2a

x = -3 ± √(9 - 4) / 2(1)

x = (-3 ± √5)/2

Question 4 :

Find all real numbers satisfying 4x − 3(2x+2 ) + 25 = 0.

Solution :

(2^2)x − 3(2x 2) + 25 = 0.

(2x)2 − 12(2x) + 25 = 0.

Let 2 =  t  ---(1)

t2 − 12t + 32 = 0

t2 − 8t - 4t + 32 = 0

t(t - 8) - 4(t - 8)  =  0

(t - 4)(t - 8)  =  0

t  =  4, t =  8

 2x  =  t2x  =  8x = 3 2x  =  t2x  =  4x = ±2

Hence the values of x are 2, -2 and 3.

Question 5 :

Solve the equation 6x4 − 5x3 − 38x2 − 5x + 6 = 0 if it is known that 1/3 is a solution.

Solution : 1/3 and 3 are solutions of the given polynomial, to find other two solutions, let us solve the quadratic equation.

6x2 + 15x + 6  =  0

6x2 + 12x + 3x + 6  =  0

6x(x + 2) + 3(x + 2)  =  0

(6x + 3)(x + 2)  =  0

x  =  -1/2 and x = -2

Hence the solutions are -1/2, -2, 3 and 1/3.  After having gone through the stuff given above, we hope that the students would have understood, "Solving Polynomial Equations with Different Powers".

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