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X + 4 4x + 7 7 14.

1.
3

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Ans. 2 =

21, X" = 5 15. (x - 1)2 = 2(x2 + 1).

Ans.

1, 2c" = - 1.

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145. Many equations of a higher degree than the second, may be reduced to the form of the second degree, and then solved. One of the most important classes of such equations consists of what are called Trinomial Equations. Such equations contain three kinds of terms, viz.: terms involving the unknown quantity to any even degree, terms involving the unknown quantity to a degree half as great, and known terms. Such, for example, as,

423
32, 2+

2x2 Every trinomial equation may be reduced to the forni of,

221 + 2px"

26

= 3.

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in the same way that equations of the second degree are reduced to the form of,

act + 270 = 9

When reduced to the required form, we may regard 201 as the unknown quantity, and then they may be solved by the same rule as is given for the solution of equations of the second degree. Having found the two values of x^, we may find the values of X, by extracting the nth root of these values. To illustrate, let it be required to solve the equation,

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This is of the required form. Writing out the values of 2), by the rule, we have,

2'3 = 2+32+4 = 8, and 23 = 2

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Whence, by extracting the cube roots of these values, we have, a' 3/8 = 2, and x'' =

- 4.

3

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This is of the required form. Writing out the values of 2%, we have,

mo"? = 1+ /3+1 -= 3, and c'!2 = 1 - 3+1 = -1. Whence, by extracting the square roots of these roots, we have,

= V3, and x" = +/- 1.

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Here, a'2 = 4+9+16 = 9, and x”? = 4–9+16 = ·-1

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= £ 3.

ac" = +/- 1.

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Extracting the cube roots of these values, we have,

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Radical equations may, also, by transformation, give rise to equations of the second degree. To illustrate, let it be required to solve the equation,

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Squaring, 42c2 + 20x = 256 - 32.0 + 202.

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1. Find two numbers whose difference is 8, and whose product is 128.

Let 2 denote the lesser number; then will x + 8 denote the greater nmber. From the conditions of the problem,

2(x + 8) = 128. Solving this equation, we find,

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Hence, the numbers are 8 and 16; also, - 16 and

- 8. Either pair of numbers will satisfy the conditions of the problem, as may be seen by trial.

In the second solution, we find the lesser number, – 16, and the greater one,

-8. Algebraically speaking, – 16 < – 8, and in general, of two negative quantities, that is algebraically the least which is numerically the greatest.

2. A person travelled 105 miles at a uniform rate. On his return, he travelled 2 miles per

hour slower, and was 6 hours longer in making the journey. How many miles did he travel per hour in the first instance ?

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