CHAPTER VIII. ON PURE EQUATIONS. 388. Equations are considered as of two kinds, called simple or pure, and adfected; each of which are differently denominated according to the dimensions of the unknown quantity. 389. If the equation, when cleared of fractions and radical signs or fractional exponents, contain only the first power of the unknown quantity, it is called a simple equation. 390. If the unknown quantity rises to the second power or square, it is called a quadratic equation. 391. If the unknown quantity rises to the third power or cube, it is called a cubic equation, &c. 392. Pure equations, in general, are those wherein only one complete power of the unknown quantity is increased. These are called pure equations of the first degree, pure quadratics, pure cubics, pure biquadratics, &c., according to the dimension of the unknown quantity. &c. Thus, xa+b is a pure equation of the first degree; x2=n2+ab is a pure quadratic; x3=a3 + a2b+c is a pure cubic ; x2=a+a3b+aca+d is a pure biquadratic; 393. Adfected equations are those wherein different powers of the unknown quantity are concerned, or are found in the same equation. These are called adfected quadratics, adfected cubics, adfected biquadratics, &c., according to the highest dimension or power of the unknown quantity. Thus, x+ax=b, is an adfected quadratic ; x2+ax2+bx=c, an adfected cubic; 3 2 x2+ax3+bx2+cx=d, an adfected biquad ratic. In like manner other adfected equations are denominated according to the highest power of the unknown quantities. I. SOLUTION OF PURE EQUATIONS OF THE FIRST 394. We having already delivered, under the denomination of Simple Equations, the methods" of resolving pure equations of the first degree, in all cases, except when the quantity is affected with radical signs or fractional exponents, in which case the following rule is to be observed. RULE. 395. If the equation contains a single radical quantity, transpose all the other terms to the contrary side; then involve each side into the power denominated by the index of the surd; from whence an equation will arise free from radical quantities, which may be resolved by the rules pointed out in Chap. III. If there are more than one radical sign over the quantity, the operation must be repeated; and if there are more than one surd quantity in the equation, let the most complex of those surds be brought by itself on one side, and then proceed as before. Ex. 1. Given √(4x+16)=12, to find the value of x. Squaring both sides of the equation, 4x+16=144; by transposition, 4x=144-16; .'.x=32. Ex. 2. Given (2x+3)+4=7, to find the value of x. By transposition, 3/(2x+3)=7-4=3; cubing both sides, 2x+3=27; by transposition, 2x=27-3; .. x=12. (12+x)=2+√x, to find the va Ex. 3. Given lue of x. By squaring, 12+x=4+4√x + x; by transposition, 8=4/x, or √x=2; .. by squaring, x=4. (∞+40)=10−√x, to find the va Ex. 4. Given fue of x. By squaring, x+40=100—20/x+x; by transposition, 20/x=60, or x=3; .. by squaring, x=9. Ex. 5. Given lue of x. (x−16)=8−√x, to find the va By squaring both sides of the equation, Ex. 6. Given √(x—a)=√xa, to find the value of x. Ex. 7. Given √5×√(x+2)=5x+2, to find the value of x. By squaring, 5x+10=5x+4/5x+4; by transposition, 6=4/5x, ..√5x=; by squaring again, 5x=,.. • to find the value of x. Multiplying both sides of the equation by ~, —ax===1, or (1—a)x=1; ••. x=; Ex. 9. Given Tue of x. √x+28_√x+38 1 1 to find the va Multiplying both sides by (x+4)×(√x+6), we have x+34√/x+168=x+42√x+152; by transposition, 16=8/x, or 2=x; .. by squaring, x=4. Multiplying both sides by (ax+b)×(31ax+5b), 3ax+2b/ax-5623ax+b/ax-263, .. by transposition, bax=362; by division, ax=3b; 962 α .. by squaring, ax=9b2, and x= Ex. 11. Given √(x+√x)−√(x−√x)= '(x) Multiply both sides of the equation by √(x+√x), 9 2 .. by transposition, x √(x) = √(x2 − x) ; 2. and dividing by √x, √x-1=√(x-1); .. by squaring, x-√x+1=x-1; ··· √x=1, 25 and by squaring, x= 16 Ex. 12. Given √(x−24)=√✓✓x-2, to find the va lue of x. Ex. 13. value of x. Given√(4α+x)=2√(b+x)−√x, to find Ans. x=49. (b-a)2 Ans. x= 2a-b Ex. 14. Given x+a+(2ax+x2)=b, to find the |