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The exponent of x in the first term is the square of its exponent in the second.

Transposing, and uniting similar terms,

[blocks in formation]

x4.

45. Find the value of x in the Equation 2x4x2+20=23.

x33; whence x= 3/3.

Transposing, and uniting similar terms,

2x4x2-3.

Dividing both sides by the coefficient of 24, we have

x2

2

x4.

Completing the square, we have

x2

2

+

x2

=

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3 12

1

4

16

Extracting the square root of both sides,

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3

2

+

[ocr errors]
[ocr errors]

16

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From this Equation we shall find

6

x2

=

hence = √√6=1√6, (237).

46. Find the value of x in the Equation

3x2-5x4

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Dividing both sides by the coefficient of x, we have

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Completing the square,

5x4

3

Extracting the square root of each side,

5 x1

6

From this Equation we shall find

4

3' 3

5x4

3

[ocr errors]

=

[ocr errors]

=

[ocr errors]

==

Transposing, and changing signs,

49

9

x=313, or

6x3

-11.

or

+

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=

+

47. Find the value of x in the Equation

6x35x+1184=0.

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[ocr errors]

By raising each of these values to the 4th power, we find

1

81

25

36

5.x-6x=1184.

Dividing both sides by the coefficient of x3,

=

1184

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[blocks in formation]

y+y=

+

48. Find the value of x in the Equation

(x+12)+(x+12)=6.

Representing the binomial (x+12) by y, we have

Completing the square, we have

1

=

4

4

y‡+y3+ 1 =6+ 1⁄2 Extracting the square root of both sides,

1

2

From this Equation we shall find

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9

25

2

49. Find the value of x in the Equation.

Transposing, and changing signs,

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y=2; hence y=16.

By restoring the binomial value of
y, we have
x+12=16; whence x=4.

25

4

(2x+6)—6—— (2x+6)*

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(2x+6)*+(2x+6)‡=6.

Completing the square, we have,

(2x+6)*+(2x+6)*+ 1 6+

4

Extracting the square root of each side,

1
(2x+6)*+ +

2

From this Equation we shall find

This Equation will give

Completing the square,

(2x+6)=2.

Raising both sides of this Equation to the 4th power,

2x+6=16.

x=5.

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50. Find the value of x in the Equation

x2+11+√x2+11=42.

This Equation may be put under the form

(x2+11)+(x2+11)=42.

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[ocr errors]
[blocks in formation]
[ocr errors]

13

2

25

169

4

51. Find the value of x in the Equation

1

(2x-4)2

2

(2x-4)+

Clearing the Equation of its fractions by multiplying its two members by the least common multiple of the denominators, which is evidently 8(2x-4), we have

8(2x-4)=(2x-4)+16.

Transposing, and changing the signs,

(2x-4)1 — 8(2x-4)2=—16. Completing the square, we have

which may

1

8

(2x-4)+-8(2x-4)2+16=0.

Extracting the square root of each side, (2x-4)2-4=0.

This Equation will give

From which we shall find

+

2x-4=2.

x=3, or 1.

52. Find the value of x in the Equation

X

4x4-4x3+ =33.

2

be put under the form

This Equation may be reduced to the form of a quadratic, thus :— The first two terms of the square root of the first side will be found to be X 2x2-x; and the remainder will be -x2+

2'

1

2

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(2x2-x).

Now, the square of the root found, and the remainder, are together equivalent to the first side of the equation; hence, we shall have

1

(2x3 — x)2 — | | (2x2—x)=33.

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