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QUADRATIC EQUATIONS.

Art. 1. A quadratic equation, or an equation of the second degree, is one which involves the second power of the unknown quantity, with or without the first power of that quantity. Every equation of the second degree involving only one unknown quantity can be reduced to one of the forms, ax2+bx+c= 0, ax2+bx=0, or ax2+c=0, in which a, b, c may be positive or negative.

If the equation be reducible to the form ax2-c=0, then ax2 = c, and æ3 =—, and the value of the unknown quantity is found by ex

a

tracting the square root of the terms on each side of the equation.

с

Taking the square root of 2a2 = 2, .. x = ± {:}', the values of x

being + {} and

a

a

{:}'

a

since the square of the same positive

and negative quantity always produces the same positive quantity.* If the equation, when reduced, assume the form x2+bx = 0, or x(x+6)=0, the values of x are x = 0 and x = —

= -b.

If the reduced equation involve both the first and second powers of the unknown quantity, as ax2+bx+c=0; when each term is divided by a, it becomes x2+

a

b

+0, and putting p and q for- and

bx с
a a

a

respectively, the equation becomes x2+px+q = 0, or x2+px = −q.

The expression apr may constitute the first two terms of the square of a binomial quantity x±a, whose square is x2±2ax+a2. And it is obvious that four times the product of the first and third terms of this expression is equal to the square of the second.

Hence if m be such a number as when added to x2+px, the expression x2+px+m becomes a complete square; then 4mx2= p2x2, and == {}; or, the quantity to be added to a2±px to make it a

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4

2

2

* As the square root of every quantity is both positive and negative, the sign equally marks both the left and right sides of an equation, when the square root is either extracted or expressed.

When the square root of 2 or x2=b (putting for to simplify the ex

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pression), the equation assumes the form ±x=±b, which gives four equalities: +x=+b3, +x=−b*; −x=+b1, -x=-bl.

Of these the first and fourth are obviously the same, as also the second and third. Hence x= ±b includes the four expressions, and it will be sufficient to mark only one side of the equation with the double sign ±.

complete square, is the square of half the coefficient of the first power of the unknown quantity, and x2±px+ is a complete square.

p2
4

By means of this property a complete quadratic equation can be reduced to a simple equation, and the values of the unknown quantity determined.

2. To solve the quadratic equation x2+px+q=0, or to find the values of x which satisfy the equation.

Taking the form x2+px+q=0, and transposing a to the other side of the equation, x2+px=-q, then adding the square of half the coefficient of x to each side of the equation, in order to make the left side a complete square, it becomes x2+px+2* extracting the square root of each side,

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—p—(p2—4q)*

2

4

are the two values of the unknown quantity of the equation

x2+px+q=0.

In the same manner, if x2+px—q = 0 be the form of the equation, then x=—P±(p2+4q)' *

2

If p2+4q be a complete square, then (p2+4q) is an integer, and the two values of x are positive or negative integers or fractions.

If p2+4q be not a complete square, then ( p2+4q)* is a surd number, and the values of x can only be approximately determined by extracting the square root of p2+4q to any number of decimal places required.

When the quadratic equation is of the form

x2+px+q=0, x = - p± (p2 - 4q).

2

If p2-4q = 0, then p2 = 4q, and the two roots of the quadratic equation are equal.

If p2 be greater than 4q, and if p2-4q be a complete square, then the two roots of the equation are rational; if p2-4q be not a complete square the two roots are irrational.

*Solve the quadratic equation x2 - 12x=13, and verify the correctness of the values of x. Here 2-12x=13. Complete the square by adding to each side of the equation the square of half the coefficient of x, then x2-12x+36=13+36=49. Extracting the square root of each side of the equation, x-6=±7, x=6±7, or 2-13 and-1.

Verification: when x=13, x2 - 12x=(13)2 - 12 × 13=169-156=13.
When x=- -1, (-1)2-12× -1=1+12=13.

And both values of x satisfy the proposed equation.

But if p2 be less than 4q, p2 - 4g is a negative quantity, the square root of which is impossible, and both the roots of the equation are impossible.

3. The quadratic equation ax2+bx+c=0 may be solved without reducing the coefficient of x to unity.

Transposing c, the equation is ax+bx-c, then multiplying each side by four times the coefficient of x, the equation becomes 4a2x2+4abx=-4ac, or (2ax)+26(2ax)=-4ac.

Next adding b2, the square of the coefficient of x, to each side, then (2ax)+26(2ax)+b2=b2-4ac, the left side of which is a complete

square.

Extracting the square root of each side, 2ax+b= ± (ba — 4ac)*,

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ac

The equation ax2+bx+c=0 may also be solved by multiplying both sides of the equation by the coefficient of x2, then a2x2+abx = or (ax)2+b(ax)=—ac, in which the unknown quantity x is changed into ax, a multiple of that quantity.

Next add to each side the square of half the coefficient of ax, then b2b2-4ac and extracting the square root,

(ax)2+b(ax)+=+

4

=

then ax+2=

4

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4. It may be remarked, that when the second power of the unknown quantity is affected with a coefficient, this method of solution will be found the most convenient in practice.

* Solve the equation 13x2 +7x=360. Multiply each side by 4 times the coefficient of x2, then 676x2 +364x=18720. Add to each side the square of 7, and 676x2 +364x+49=18720 +49-18769. Extracting the square root, 26x+7=137, and 26x=-7±137=130 and - 144,

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Here both values of a satisfy the equation.

Solve the equation 7x-4x=3, and verify the correctness of the values of x. Multiply each side of the equation by 7, then (7x)2 — 4(7x)=21.

Next complete the square, (7x)2 - 4(7x) + 4 =21+4=25.

Extract the square root, 7x-2=±5, and 7x=2±5-7 and -3,

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In the solution of equations of the second degree, certain anomalies arise, besides those which occur in the solution of equations of the first degree. It is possible that either one or both of the values of the unknown quantity obtained do not satisfy the proposed equation. This anomaly arises when a term in the proposed equation containing the unknown quantity is affected by the square root.

Every quantity has two equal square roots, one positive and the other negative, and if an expression of the form +(mx+n) occur in an equation, the square of this quantity is me+n, the same as the square of (mx+n)1.

-

Any proposed equation involving this expression, when reduced may be pr+q+(mx+n) or px+q=−(mx+n), and when each side of these equations is squared, they assume the same form

(px+q)2 = mx+n in both cases.

Here an ambiguity is introduced in the course of the solution, and in the equations so reduced the distinction does not appear which existed in the proposed equations.

Hence it follows that a new condition is introduced, which will be manifest in the result, so that one or both values of the unknown quantity obtained may not satisfy the proposed equation. The roots so introduced are called roots of solution, as distinguished from the proper roots which always satisfy the equation.*

5. A quadratic equation cannot have more than two values of the unknown quantity.

Let a, b be the values of x in the equation a+px+q=0, and, if possible, let c be a third value of x.

Solve the equation 2x+(7x+11)=9.

Here (7x+11)=9-2x by transposing 2x, next 7x+11=81-36x+4x2, by squaring each side of the equation, and 4x2 - 43x=-70 by transposition. Multiply each side by 4, and (4x)2 - 43(4x)=-280.

Complete the square on the left side by adding to each side the square of half the coefficient of the first power of the unknown quantity,

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If these two values of x satisfy the proposed equation, they are the proper roots

of the equation.

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which ought to be 9, and x=" does not satisfy the equation, and is a root of

solution.

Let x=2, then 2x+(7x+11)1=4+ (14+11)1=4+(25)1=4+5=9.

Hence x=2 does satisfy the proposed equation, and is a true root of it.

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