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Completing the square by adding to each side

b

72 x2+ -x+

4a?

b 62-4ac that is,

x +

;

2a 4a? extracting the square root,

b IV(62 4ac). x +

-;
2a

2a
-5+(b2 - 4ac)

2a Note. The student will observe that b, the first term of the numerator of the fraction, is the coefficient of x in the original equation with its sign changed, and that 4ac, under the radical, is plus or minus according as the signs of a and c in the original equation are like or unlike.

214. Instead of going through the process of completing the square in each particular example, we may now make use of this general formula, adapting it to the case in question by substituting the values of a, b, c.

Example. Solve 5x2 + 11x – 12=0.
Here a=5, b=11, c= - 12.

-11 V(11)2–4.5(-12)

10
-11 + 1361 -11+19 4
10

or - 3.
10 5'

-6+(12-4ac) 215. In the result

2a it must be remembered that the expression (12 4ac) is the square root of the compound quantity b2 - 4ac, taken as a whole. We cannot simplify the solution unless we know the numerical values of a, b, c. It may sometimes happen that these values do not make b2 — 4ac a perfect square. In such a case the exact numerical solution of the equation cannot be determined.

.. 20 =

X =

Example 1. Solve 5x2 – 15x + 11=0.

15€ (-15)2 – 4.5.11 We have

2.5 15 € 15

10 Now

V5=2-236 approximately.
15 +2.236

-=1.7236, or 1.2764.

10 These solutions are correct only to four places of decimals, and neither of them will be found to exactly satisfy the equation.

Unless the numerical values of the unknown quantity are required it is usual to leave the roots in the form

15+ V5 15 – V5
10

10

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Example 2. Solve 22 - 3x +5=0.

3+1 - 3)2 – 4.1.5 We have

2
3*19 - 20

2
3+N-11

2 But – 11 has no square root exact or approximate [Art. 178]; so that no real value of x can be found to satisfy the equation. In such a case the roots are said to be imaginary or impossible.

SOLUTION BY FACTORING.

2.

.......

216. The following method will sometimes be found shorter than either of those already given. Consider the equation ++

7
x2 + x=

3
Clearing of fractions, 3x2 +7-6=0

..(1); by resolving the left-hand side into factors we have

(3.x - 2) (x+3)=0. Now if either of the factors 3x – 2, x+3 be zero their product is

Hence the quadratic equation is satisfied by either of the suppositions

3x – 2=0, or x+3=0.

2 Thus the roots are

-3. 3

zero.

It appears from this that when a quadratic equation has been simplified and brought to the form of equation (1), its solution can always be readily obtained if the expression on the left-hand side can be resolved into factors. Each of these factors equated to zero gives a simple equation, and a corresponding root of the quadratic.

Example 1. Solve 2.02 – ax + 2bx=ab.
Transposing, so as to have all the terms on one side of the equation,

we have

Now

2x2 - ax + 2bx ab=0.
2.02 – ax + 2bx ab=x (2.c – a) +6 (2.c – a)

= (2x – a) (x+b).
(2x – a) (a+b)-=0;

2.c - a=0, or x+b=0.

Therefore whence

a

(1).

or - b.

2' Example 2. Solve 2(x2 - )=3 (x-4). We have

2x2 – 12=3x – 12; that is,

2.2 = 3.0 Transposing,

2.c – 3x=0,
2 (22C – 3)=0.
c. x=0, or 2.c – 3=0.

3 Thus the roots are 0,

2•

NOTE. In equation (1) above we might have divided both sides

3 by x and obtained the simple equation 2x=3, whence a= which is

2' one of the solutions of the given equation. But the student must be particularly careful to notice that whenever an x is removed by division from every term of an equation it must not be neglected, since the equation is satisfied by x=0, which is therefore one of the roots.

217. There are some equations which are not really quadratics, but which may be solved by the methods explained in this chapter.

Example 1. Solve 24 – 13x2 +36=0.
By resolution into factors, (2c2 – 9) (oc? – 4)=0;

.:. x2 – 9=0, or x2 - 4=0; that is,

x==9, or 4, and

x=+3, or +2.

EXAMPLES XXVI. c.

Solve by the aid of the formula in Art. 213: 1. 3x2=15 - 4x. 2. 2x2 +7x=15. 3. 2.x* +7 – 9x=0. 4. X=3x+5. 5. 502 +4+21x=0. 6. x2 +11=7x. 7. 8.x2=x+7.

8. 522=178 - 10. 9. 35+92 – 2x2=0.

10. 3.-=x+1. 11. 3x2 + 5x=2.

12. 2x2 +5% - 33=0. Solve by resolution into factors : 13. 6.x2= 7+2.

14. 21 +8x"=26x. 15. 26x – 21+11x2=0. 16. 5x2 +263 +24=0.

4 17. 4.x2 +3.

18. 22 – 2=15%. 15 19. 702=28 - 96x.

20. - 96x2=4x + 15. 21. 25x2=5.x+6.

22. 35 - 4x=4x2. 23. 12x2 – llax=36a2.

24. 12.x2 + 360o=43ax. 25. 3562=9.x2 +66x.

26. 36x2 – 3562=126%. 27. x2 – 2ax +4ab=2bx. 28. x2 – 2ax +8x=16a. 29. 3x2 – 2ax - bx=0,

30. ax2+2x=bx. Solve as explained in Art. 217 : 31. 4=5.22 – x4.

32. X4+36=13x2. 33. 26+7x3=8.

34. 206 – 19.23=216.

CHAPTER XXVII.

EQUATIONS IN QUADRATIC FORM.

218. An equation in the form ax2n +bx"=c, n being a positive or negative integer or fraction is in quadratic form. Thus 24 +4x2=117, x*+7x3=44, and x-}+x-1=a are equations in quadratic form.

In Art. 217 we solved an equation of this character by factoring. We now give a few more examples showing that the ordinary rules for quadratic equations are applicable to those in quadratic form.

Example 1. Solve 24 – 13x2= -36.
Using the formula of [Art. 213].

13+ V(13)2 -- 4(36)
22

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