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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;
extracting the root, x+1=15;
x=15—1=4, or -6.

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3. Given y—ya+5=8+y3, to find the value of y.

By transposition, y—2y+—3;
completing the square, y—2yž+1=4;

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;

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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+

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squaring each side, 2x+4=16;
by transposition, 2x=12; .. x=6.

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multiplying this equation by 6-x,
12x2+60x+48-146x-25x2+24;

by transposition, 37x2-86x=-24;

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Multiplying the numerator and denominator of

the fractional part by x+x2-4,

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Taking the positive quantity, and multiplying by 2,

x+√/x2-4=4x-6;

by transposition, and squaring each side,
x2-4-9x2-36x+36 ;

by transposition, 8x-36x=-40;

dividing by 8, and completing the square,

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squaring each side, x2-4-25x2-60x+36;
by transposition, 24x2-60x=-40;

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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.

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Let the positive value be taken, then by transpo

sition,

2x2-8x+15=1+2=3;

squaring each side, 2x2-8x+15=9;
by transposition, 2x2-8x=-6.
dividing by 2, x2-4x=-3;

completing the square, x2-4x+4=1;
extracting the root, x-2=±l;

.. 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;
by transposition, and dividing by 2,

x2-4x=-7;

completing the square, 4x+4=-3;
extracting the root, x-2=±√−3;
by transposition, x=2±√−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)+—

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extracting the root, x2-x-2=;

59

·· x2 - 2 = ± 2 + 30, or -29.

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