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co-efficient of its second term ; then extract the square root of each side of the equation ; and there will arise a simple equation, by which the value of the unknown quantity may be determined.
Examples 1. Given x + 6x=40 to find the value of x.
Solu.-- We know that x? + 6x +9 is the square of x+3; for (x+3)=x' +6r+9.
If, therefore, we add 9 to each side of the equation, there will arise a new equation x2 + 6x+9=40+9=49, of which one side is the square of x+3, and the other side is the square of 7.
By extracting the square root of each side of this last equation,
W x* +6r+9=v49; or 3+3=7;
Therefore x=7—3=4, Ans.
and we have z?-10x+25=11+25=36. Extracting the square root, we have r-5= 36=6;
Therefore, x=6+5=ll. 3. Let x2 + 2ax=l, to find x. Solu.—To each side of the equation add a® ;
Then xo + 2axta’=b+a';
But x + 2ax+u is the square of x+a;
Hence, x+a=v0+a®; and .. x=vota' Fa Note.-1. It will be observed, on reviewing these three examples, that, in each case, the quantity added to both sides of the equation was the square of half the co-efficient of x.
2. And in performing operations of this description, the learner should collect, that, after completing the square according to the rule, the squ root of that side of the equation containing the unknown quantity, is always , with half the co-efficient of the second term anne.red with its proper sig
Thus, the square root of x2 + 6x + 9 was x + 3; that of se – 10.0+ 25 was x—5; and that of x+2ax +ao was za ta.
Examples for Practice. 4. Given x® +87=65, to find t........
5. 5. Let 22—4x=45, to find x....
Ans. x =
9. 6. Let x2 +12x=108, to find x.. .. .. .. Ans. x = 6. 7. Let r — 14x=51, to find x..... Ans. x =
17. 8. Given to +6bx=c?, to find x... Ans. x=vc +90-36.
3 1369 37 Extract the square root, and x
64 8 37 3 40 Therefore, is +
8 8 8 3. Given 2ax' bx=c, to find x.
12 Then, add the square of and x2 2+
b Bac +12v vac+12
Examples for Practice. 4. Given 3x4 + 2x=161, to find X....... Ans. x 7. 5. Let 23% – 5x=117, to find x..... Ans. x= 9. 6. Given 3x? - 2r=280, to find x....... Ans. x=10 7. Let 7x -20r= 32, to find x.......
Ans, x= 4. 3. Given 52% + 4x=273, to find x....... Ans. r= 7.
✓ Aac+7-6 9. Let ax' + Wr=c, to find
(71.) One value only, of the unknown quantity, hath been found in the preceding questions; but every quadratic equation contains two values of the unknown quantity, arising from the following circumstance.
Since – ax – a gives + a?, as well as tax ta; when we take the square root of + a', we may obtain for a result, either + a,
- a; consequently, in extracting the square root of each side of the equation, after the square has been completed, there will arise two simple equations to determine the value of the unknown quantity.
(72.) For Example. Let x2 + Px=Q, represent a quadratic equation in its most general form, wherein P and Q are any numbers integral or fractional, positive or negative.
P Solu. To each side of the equation, add the square of
2' P2 P? P2 +4Q. And it becomes x? +Pr+
P2 +4Q tv P+42 square root,
P -VP+4Q Hence, x+ , and x +
are two 2. 2
2 simple equations for determining the value of x;
+P+4Q-P From which it appears, that r=
VP: +4Q-P -sand 2
2 Note.- If Q be a negatire quantity, and Pa be less than 4Q, then the quantity Po + 4 Q is negative; and because the square root of a negative quantity hath no real representation, the two values of the unkuown quantity, containing the radical ^p + 4Q, are said to be imaginary or impossible.
Eramples. 1. Let x? + Ox=112, to find the two values of x. Solu. First, add 9 to each side, then xo +6r+9=112+9=121;
Therefore, extracting the square root, x+3=+V 121=+11; and, consequently, r=+11—3=8, or 14 Ans.
2. Given x2 – 16r=-63, to find, in this equation, the two values of x.
Solu. Add 64 to each side, then re-16x + 64=64–63=1; consequently, x-8=tvī=+1, or x=8+1=9 or 7.
3. Let vé + 8x=-31, to find the two values of x in this equation.
4. In the equation ? 2cx=ato find the two values of r.
Examples for Practice. 5. Let ro - 2x=15, to find the two values of x... Ans. x=5 or-3. 6. Given r? - 16x=-15.....
Ans. x=15 or 1. 7. Let + Or=135..
Ans. x=9 or -15. 8. Given x-21= -2
Ans. x=l+v-. 9. Letra +12x=-50..
Ans. r=-6+- 14. 10. Given r* - Bax=30°..
.. Ans. x=40+16+31%. (73). Every equation, in which the unknown quantity is found in only two terms, and the index of the highest power is double the index of the lowest, may be reduced to the form of a quadratic equation, and solved according to the foregoing rules.
(74). For Example. Let x2 +Px"=Q be any general equation.
Solu. Let x"=y, substitute now these values for 22n and mo, and then rry; the given equation becomes y? + Py=Q.
p? P2 +4Q
+vP2 + 4Q-P Then yo+Pg+
; conse4 4
2 quently, since x"=y, x must be equal to y;
+P+4Q-P Therefore, r=
Examples. 1. Let 3* - 6x=27, to find the values of x in this equation. Solu. Let x'=y, whence the equation becomes yo -6y=27; then x'=yo;
But, because x'=y, r=+vy=+g, or tv - 3=+3, or +3. 2. Find the values of x in the equation 20—2x=48.
Solu. Let r=y, } therefore, by substituting ye - 2y=48. then r=yo;
But y - 1=+ v49=+7; hence, y=+7+1=8 or-6; consequently, r=y=y, or 4-6=2, or 5-6.
Examples for Practice 3. Find the values of x in the equation x* +4xʻ=12
Ans. x=+vž, or + v6 4. Find the values of x in the equation x® —8x=513.
Ans. x=3 or 31 – 19. 5. Find the values of x in the equation xen - 4x"=10.
Miscellaneous Examples of Quadratic Equations.
11= to find the values of r.
) Solu. First, then, 11
dividing by 4. 4
7 2 2. Given the equation + 5, to find the values of r x+1
7 2 Solu. First, multiply by x+1, and the equation +
2r +2 Multiply, again, by x and 7+
= 5x +5,
4 4 2 4 14 Completing the square re-+*+ +
25 5 25 25 2
14 +v14 Extract, now, the square root, and x +
25 5 +74+2 Therefore, I =
=1.14, or — - .34. 5