**Solving Polynomials with Rational Powers :**

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

**Question 1 :**

Solve 8 x^{3/2n }- 8 x^{-3/2n} = 63

**Solution :**

8 x^{3/2n }- 8 x^{-3/2n} = 63

8 x^{3/2n }- 8 (1/x^{3/2n}) = 63

x^{3/2n }- (1/x^{3/2n}) = (63/8)

Let t = x^{3/2n}

t - (1/t) = 63/8

8(t^{2} - 1) = 63t

8t^{2 }- 63t - 8 = 0

8t^{2 }- 64t + t - 8 = 0

8t(t - 8) + 1(t - 8) = 0

(8t + 1)(t - 8) = 0

t = -1/8, t = 8

x^{3/2n } = 8

Raise 2/3 as power for both sides

(x^{3/2n})^2/3 = (2^{3})^2/3

x^{1/n} = 2^{2}

x^{1/n} = 4

x = 4^{n}

**Question 2 :**

Solve :

**Solution :**

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

t = -b ± √(b^{2} - 4ac) / 2a

t = [(b^{2} + 6a^{2}) ± √(b^{2} + 6a^{2})^{2} - 4(2ab)(3ab)]/2(2ab)

t = [(b^{2} + 6a^{2}) ± √(b^{4} + 36a^{4 }+ 12a^{2}b^{2} - 24a^{2}b^{2})/4ab

t = [(b^{2} + 6a^{2}) ± √(b^{2}-6a^{2})^{2}]/4ab

t = [(b^{2} + 6a^{2}) ± (b^{2}-6a^{2})]/4ab

t = [(b^{2} + 6a^{2}) + (b^{2}-6a^{2})]/4ab

t = (b t = 2b t = b/2a √(x/a) = t x/a = (b/2a) x = ab x = b |
t = (b t = 12a t = 3a/b √(x/a) = t x/a = (3a/b) x = 9a |

**Question 3 :**

Solve the equations

(i) 6x^{4} − 35x^{3} + 62x^{2} − 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.

6x^{2} - 5x + 1 = 0

6x^{2} - 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) x^{4} + 3x^{3} - 3x - 1

1 and -1 are the solutions of the given polynomial, to find other two solutions, let us solve the quadratic equation

x^{2} + 3x + 1 = 0

x = -b ± √(b^{2} - 4ac) / 2a

x = -3 ± √(9 - 4) / 2(1)

x = (-3 ± √5)/2

**Question 4 :**

Find all real numbers satisfying 4^{x} − 3(2^{x+2} ) + 2^{5} = 0.

**Solution :**

(2^2)^{x} − 3(2^{x} 2^{2 }) + 2^{5} = 0.

(2^{x})^{2} − 12(2^{x}) + 2^{5} = 0.

Let 2^{x } = t ---(1)

t^{2} − 12t + 32 = 0

t^{2} − 8t - 4t + 32 = 0

t(t - 8) - 4(t - 8) = 0

(t - 4)(t - 8) = 0

t = 4, t = 8

2 2 x = 3 |
2 2 x = ±2 |

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

**Question 5 :**

Solve the equation 6x^{4} − 5x^{3} − 38x^{2} − 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.

6x^{2} + 15x + 6 = 0

6x^{2} + 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".

Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.

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