Adfected Quadratics of one unknown quantity with their solutions. 1. Given 3x2+6x+6=78, to find the values of x. By transposition, 3x2+6x=78—6—72; dividing by 3, x2+2x=24; completing the square, x2+2x+1=25; 3. Given y—ya+5=8+y3, to find the value of y. By transposition, y—2y+—3; extracting the root, y-1=2; by transposition, y=3; .. y=9. 4. Given x3-20189, to find the values of x. completing the square, x3-20x+100=289; extracting the root, x2-10-17; 3 by transposition, x=17+10=27, or -7; x=273=9, or 73 5. Given 2x+4=2x+4+12, to find the values of x. By transposition, 2x+4-2x+4=12; completing the square, 2x+4\/2x+4+ squaring each side, 2x+4=16; multiplying this equation by 6-x, by transposition, 37x2-86x=-24; Multiplying the numerator and denominator of the fractional part by x+x2-4, Taking the positive quantity, and multiplying by 2, x+√/x2-4=4x-6; by transposition, and squaring each side, by transposition, 8x-36x=-40; dividing by 8, and completing the square, squaring each side, x2-4-25x2-60x+36; 8. Given 8x-2x2+44/2x2-8x+15=18, to find the values of x. Subtracting 15 from each side of the equation, 8x-2x2-15+4/2x2-8x+15=3; changing the signs, 2x2-8x+15—4、/2x2—8x+15 -3; completing the square, 2x2-8x+15—4、√√2x2—8x+15 +4=4-3=1; extracting the root, 2x2-8x+15—2=±1. Let the positive value be taken, then by transpo sition, 2x2-8x+15=1+2=3; squaring each side, 2x2-8x+15=9; completing the square, x2-4x+4=1; .. by transposition, x=1+2=3, or 1. But if the negative value be taken, viz. /2x2-8x+15-2=-1; by transposition, 2x2-8x+15=1; squaring both sides, 2x2-8x+15=1; x2-4x=-7; completing the square, 4x+4=-3; 9. Given a1-2x3+x=870, to find the values of a. Here, adding and subtracting x2, and the equation becomes a1——2x3+x2x2+x=870; (x2—x)2 —(x2 —x)=870; completing the square, (2o—x)2—(x2—x)+— 59 extracting the root, x2-x-2=; 59 ·· x2 - 2 = ± 2 + 30, or -29. |