SOLVING POLYNOMIAL AND NON POLYNOMIAL EQUATIONS

Example 1 :

Solve : 

8x3/2n - 8x-3/2n = 63

Solution :

8x3/2n - 8x-3/2n = 63

8x3/2n - 8(1/x3/2n) = 63

Divide both sides by 8. 

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 and t = 8

When t = -1/8, 

x3/2n = -1/8

Take power 2n/3 on both sides. 

(x3/2n)2n/3 = (-1/8)2n/3

x = (-1/8)2n/3

x = [(-1/2)3]2n/3

x = (-1/2)2n

x = [(-1/2)2]n

x = (1/4)n

x = 1/4n

But x = 1/4n will not satisfy the given equation. So, we can ignore this solution. 

When t = 8, 

x3/2n = 8

Take power 2n/3 on both sides. 

(x3/2n)2n/3 = 82n/3

x = 82n/3

x = (23)2n/3

x = 22n

x = (22)n

x = 4n

The possible solution is x = 4n.

Example 2 :

Solve :

2√(x/a) + 3√(a/x) = b/a + 6a/b

Solution :

Let t = √(x/a). Then, 1/t = √(a/x). 

2√(x/a) + 3√(a/x) = b/a + 6a/b

2t + 3(1/t) = b/a + 6a/b

2t + 3/t = b/a + 6a/b

Multiply both sides by abt. 

2t2ab + 3ab = b2t + 6a2t

2t2ab + 3ab = (b2 + 6a2)t

2t2ab - (b2 + 6a2)t - 3ab = 0

Solve the above quadratic equation in the variable 't' using quadratic formula. 

t = [-b ± √(b2 - 4ac)]/2a

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

t = [(b2 + 6a2) ± √(b4 + 36a+ 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)/4ab

t = 2b2/4ab

t = b/2a

√(x/a) = t

x/a = (b/2a)2

x = ab2/4a2

x = b2/4a

t = (b2+6a2)-(b2-6a2)/4ab

t = 12a2/4ab

t = 3a/b

√(x/a) = t

x/a = (3a/b)2

x = 9a3/b2

Example 3 :

Solve : 

6x4 - 35x3 + 62x2 - 35x + 6 = 0

Solution :

This is a polynomial equation of degree 4. So, we will have four zeros. 

By trial and error, we can check the values 1 or -1 or 2 or -2...... as zeros for the above equation using synthetic division. 

6x4 - 35x3 + 62x2 - 35x + 6 = 0

2 and 3 are the two zeros of the given equation. To get the other two zeros, solve the resulting quadratic equation in the above synthetic division. 

6x2 - 5x + 1 = 0

6x2 - 2x - 3x + 1 = 0

2x(3x - 1) - 1(3x - 1) = 0

(2x - 1)(3x - 1) = 0

x = 1/2 and 1/3

The four zeros are 2, 3, 1/2 and 1/3.

Example 4 :

Solve :

x4 + 3x3 - 3x - 1 = 0

Solution : 

This is a polynomial equation of degree 4. So, we will have four zeros. 

By trial and error, we can check the values 1 or -1 or 2 or -2...... as zeros for the above equation using synthetic division. 

x4 + 3x3 - 3x - 1 = 0

1 and -1 are the two zeros of the given equation. To get the other two zeros, solve the resulting quadratic equation in the above synthetic division. 

x2 + 3x + 1 = 0

x = [-b ± √(b2 - 4ac)]/2a

x = [-3 ± √(9 - 4)]/2(1)

x = (-3 ± √5)/2

The four zeros are -1, 1, (-3 + √5)/2 and (-3 - √5)/2.

Example 5 :

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

Solution :

4x - 3(2x + 2 ) + 25 = 0

(22)x - 3(2x22) + 32 = 0

(2x)2 - 12(2x) + 32 = 0

Let y = 2x.

y2 - 12y + 32 = 0

y2 - 8y - 4y + 32 = 0

y(y - 8) - 4(y - 8) = 0

(y - 4)(y - 8) = 0

y = 4 and 8

y = 4

2= 4

2= 22

x = 2

2= 8

2= 23

x = 3

The two zeros are 2 and 3.

Example 6 :

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

Solution :

Given : 1/3 is a solution.

1/3 and 3 are the two zeros of the given equation. To get the other two zeros, solve the resulting quadratic equation in the above synthetic division. 

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 -2

The solutions are 1/3, 3, -1/2 and -2. 

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