AN AFFECTED QUADRATIC EQUATION is that, which involves the square of the unknown quantity, together with the product, that arises from multiplying it by some known quantity.

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Thus, ax is a simple quadratic equation,

And ax+bxc is an affected quadratic equation.

The rule for a simple quadratic equation has been given already.

All affected quadratic equations fall under the three following forms.

The rule for finding the value of x, in each of these equations, is as follows;

1. Transpose all the terms, that involve the unknowa quantity, to one side of the equation, and the known terms to the other side, and let them be ranged according to their dimensions.

2. When

* The square root of any quantity may be either + or and therefore all quadratic equations admit of two solutions. Thus, the square root of +n2 is +n, or -; for either +nx +n, or n×—n is equal to +n. So in the first form, where

2 --?

or - √ b+ since either of them being multiplied by itself

And this ambiguity is expressed by writing the

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2. When the square of the unknown quantity has any coefficient prefixed to it, let all the rest of the terms be divided by that coefficient.

3. Add

be negative, because it is composed of two negative terms,

Therefore, when x+axb, we shall have x+√o+

2

for the affirmative value of x, and x——√ b+.

for the negative value of x,

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In the second form, where xb+

a

a

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value, viz. x=+√ b+ + is always affirmative, since

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2

it is composed of two affirmative terms.

4

2

a + will always be negative; for since!

3. Add the square of half the coefficient of the second term to both fides of the equation, and that side, which involves the unknown quantity, will then be a complete square.

4. Extract

when x-axb, we shall have x=+√ b+

affirmative value of x2 and

being composed of two affirmative terms. The second value,

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a

—b, and consequently —√2 -b+will always be

4

2

an affirmative quantity. Therefore, when x2-axb, we shall

have

4. Extract the square root from both sides of the equa❤ tion, and the value of the unknown quantity will be de termined, as required.

I.

NOTE 1. The square root of one side of the equation is always equal to the unknown quantity, with half the co efficient of the second term subjoined to it.

NOTE 2. All equations, wherein there are two terms involving the unknown quantity, and the index of one is just double that of the other, are solved like quadratics by completing the square.

Thus, xxb, or x"+ax2=b, are the same as quadratics, and the value of the unknown quantity may be determined accordingly.

the proposed question will be impossible. For, since the square of any quantity (whether that quantity be affirmative or negative) is always affirmative, the square root of a negative quantity is impossible, and cannot be assigned. But if be greater than, 4

then

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-b is a negative quantity; and consequently

a

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is impossible, or only imaginary, when is less than b; and

therefore in that case x= +b is also impossible or

Imaginary.

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EXAMPLES.

1. Given x+4x=140; to find .

First, x+4x+4=140+4 144 by completing the

square;

Then x+4x+4=144 by extracting the root; Or x+2=12,

And therefore x12—2—10.

2. Given x2-6x+8=80; to find x.

First, x-6x= 80-872 by transposition; Then x2-6x+9=72+9=81 by completing the square ; And x-3819 by extracting the root;

Therefore

9-312

3. Given 2x+8x-20-70; to find x.

First, 2x+8x=70+20=90 by transposition.
Then x+4x=45 by dividing by 2,

And x2+4x+4=49 by completing the square ;
Whence +2=√ 497 by extracting the root,
And consequently 7—2—5.

4. Given 3x-3x+6=5; to find x.
Here x-x+2=1 by dividing by 3,
And x-x=13-2 by transposition;

Also x2+12+1=3 by completing the x*

square,