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Taking the square root of both members of the last equation giyes

x+3=+5. Therefore, we must either have x+3=5 or x+3=-5.

Solving the last two equations separately gives as the desired solutions x=2 and x=-8. Ans.

CHECK. Substituting 2 for x in the first member of the given equation gives 22+6x2 or 4+12 which is 16 as required. Similarly, when x=-8 the first member becomes (-8)2+6X(-8), or 64-48, which is 16 as required.

EXAMPLE 2. Solve x2 – 8 x+14=0.
Solution. Transposing, we find

2–8 x=-14.
Completing the square by adding 16 to both sides gives

x2–8 x+16=2, or (x-4)2=2. Taking the square root of both members,

x-4==V2. Solving the last two equations gives

x=4+V2 and x=4-V2. Ans. CHECK. With x=4+V2, the first member of the given equation becomes (4+V2)2–8(4+12)+14. By Formula V, p. 101, this may be written

(16+8V2+2) –8(4+V2)+14, or 16+8V2+2–32–8V2+14. Here the 8V2 and 48V2 cancel, while the rest of the expression (namely, 16+2–32+14) reduces to 0. Whence, if x=4+V2 the first member reduces to 0, as required.

Likewise, when x has its other value, namely 4-V2, the first member of the given equation may be shown at once to reduce to O as required.

Note. Since the solutions obtained above for Example 2, namely x=4+V2 and x=4-V2, contain the surd V2, they cannot be expressed exactly (see § 135), but we can express them approximately. Thus, the Tables give V2=1.41421, so that the two solutions are approximately 4+1.41421 and 4-1.41421, which reduce to 5.41421 and 2.58579. Ans.

EXAMPLE 3. Solve 3 x2 +8 x=15.
Solution. Dividing through by 3, we find

x2+$ x=5. Completing the square by adding (1)2, or (4) to both members gives

x2+{ x+(1)2=5+48 = 1

or,

(x+4)2=1. Taking the square root of both members gives

x+f= =VI= +4V61. Therefore the two solutions are

x=-4+1V61, and x=-1-1V61. Ans. Note. These two answers may be written together in the condensed form }(-4=V61). By looking up the value of v61 in the Tables, the value of each answer may be determined approximately, as explained in the Note to Example 2.

155. Summary of Results and Rule. It is now to be carefully observed that in each of the three examples worked in § 154 the first step consisted in reducing the given equation to the type form

x2+px=9, where p and q are certain known numbers.

Thus, in Example 3 we first put the equation in the form x2+x= 5. Here prf and q=5.

The next step was to complete the square by adding to both members the square of half the coefficient of x; that is, we added (p/2)2 to both members.

Thus, in Example 3, where p=8/3, we have p/2=4/3; hence the number added to both sides to complete the square was (4/3)2.

After this, the equation was always such that we could extract the square root of the left member. When we did so and equated (placed equal to each other) the results, we had two simple equations from which the two solutions could be found at once.

This may now be summarized into the following rule.

RULE FOR SOLVING ANY QUADRATIC BY COMPLETING THE SQUARE. 1. Reduce the equation to the form

x2+px =q. 2. Complete the square by adding (p/2)2 to each member.

3. Extract the square root of both members of the equation thus obtained and equate the results. This yields two simple equations from which the two desired values of x may be found.

EXERCISES Solve the following equations, and check your answers in the first five. 1. x2 +4 x=12.

14. (3 x-2)2=6 x +11. 2. x2–6 x=16.

16. x + 15 – 16. 3. x2 — 20 x=21. 4. 22–18 x= -80. 16. 922–3 x=-1. 5. x2 +3 x=1. 6. x2 – 7 x= -10.

17. x2 – 24 = 10. 7. x=1-«2. 8. 22– fx=1.

18. 72 = x. 9. 3 x2 —4 x=1.

2—3 _ 6-2 10. 5 x2 +25 x= -9.

x+5 x-2 11. 6 x2=3 x+45.

20. –2 a?=ax — . 12. x2—4 x=6(x+4).

x=2 a, x=-a. Ans. 13. 2 x(+4)=42.

21. 12 x2 – 3 ax= 3 a’. For further exercises on this topic, see the review exercises, p. 285, and Appendix, p. 310.

156. Quadratics of Special Form. Suppose we wish to solve the quadratic 4 x2+4 x=15. Here the coefficient of x2 is 4, or 22, and is therefore a perfect square. In such a case we do not need to first divide the equation through by 4, as the Rule of § 155 would require, but we can complete the square immediately in the equation as it stands. Thus, by adding 1 to both members we have

4x2+4 x+1=16, which is the same as (2 x+1)2 = 16. Extracting the square root of both members of the last equation gives 2 x+1= +4, so the two solutions are those of the simple equations 2 x+1=4 and 2 x+1= -4; that is, they are x=3/2 and x=-5/2. Ans.

All quadratics such as the one just mentioned are of the type form

n2x2+bx=k, where n, b, and k are certain determinate numbers, and we can immediately complete the square by adding (b/2 n)2 to both members.

Thus, in the quadratic 25 x2–30 x=72 we have n=5 (because na=25) and we have b=-30, k=72. So the square can be immediately completed by adding (-30/10)2= (-3)2=9 to both members. This gives 25 x2—30 x+9=81, or (5 x-3)2=81, from which we find the solutions x= and x=-6. This amounts to the same thing as the process of $ 155, as can be seen in this example by first dividing both sides of the given equation by 25.

It should be observed that if n=1 the type form we have just been considering reduces to the one employed in § 155.

EXERCISES Solve the following by the method of completing the square explained in § 156. Check your answers in the first five. 1. 4 x2 +4 x=8. 2. 4x2 — 4 x= 15. 3. 9x2 – 12 x=5.

4. 16 x2 +8 x-1=0. 5. 16 x2 – 10 x=- 41.

11-1= V2). Ans. 6. 36 x2 +6 x=. 7. 2 x2+6 x= 1.

[Hint. First multiply both members by 2 so as to get a perfect square for the coefficient of x2.]

8. 3 x2 – 2 x=5.
[Hint. Multiply both members by 3.j
9. 5 x2 — 20 x+14=0. 10. 8x2— 5 x – 1=2.

*157. Solution by the Hindu Method. A simple way, preferred by many, for completing the square in any quadratic is the one called the “Hindu Method.” It consists of two steps:

1. Multiply both members by four times the coefficient of x2.

2. Add to both members of the new equation the square of the original coefficient of x.

EXAMPLE. Solve 3 x2–2 x=1.

Multiplying through by four times the coefficient of x2, that is by 12, gives

36 x2–24 x=12. Adding the square of the original coefficient of a to both sides, that is, adding (-2)2 or 4 to both sides, gives

36 x2— 24 x+4=16. The first member is now a perfect square, being equal to (6 x—2). Therefore, upon extracting square roots, we obtain

6 x—2=4, and 6 x-2=-4. Solving the last two equations, gives x=1, and x=-1. Ans.

EXERCISES Solve each of the following equations by any method. 1. x2+10 x+21=0.

4. 22–4 x=117. 2. x2–5 x=24.

6. 8x=x2–180. 3. 2 x2+7 x=60.

6. 5x2–3 x-2=0.

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