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# publicKhan/khan-exercises

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 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103         Solving quadratics by completing the square 1

randRangeNonZero( -10, 10 )                    randRange( -5, 5 ) * 2 + ( X1 % 2 )                    ( X1 + X2 ) * -1                    X1 * X2                    new Polynomial( 0, 2, [C, B, 1], "x" )                    POLY.text()                    \$._("or")

Complete the square to solve for x.

POLY_TEXT = 0

B/2                        C * -1 + pow( B / 2, 2 )                        X1                        X2

B/2                        C * -1 + pow( B / 2, 2 )                        X2                        X1

Completed Square:
(x + {} )^2 = {}

Solution:
x = {}\quad\text{OR}\quad x = {}

X1

Completed Square:
(x + {}-X1 )^2 = {} 0

Solution:
x = \quadX1

Begin by moving the constant term to the right side of the equation.

x^2 + Bx = C * -1

We complete the square by taking half of the coefficient of our x term, squaring it, and adding it to both sides of the equation. Since the coefficient of our x term is B, half of it would be B / 2, and squaring it gives us \color{blue}{pow( B / 2, 2 )}.

x^2 + Bx \color{blue}{ + pow( B / 2, 2 )} = C * -1 \color{blue}{ + pow( B / 2, 2 )}

We can now rewrite the left side of the equation as a squared term.

( x + B / 2 )^2 = C * -1 + pow( B / 2, 2 )

The left side of the equation is already a perfect square trinomial. The coefficient of our x term is B, half of it is B / 2, and squaring it gives us \color{blue}{pow( B / 2, 2 )}, our constant term.

Thus, we can rewrite the left side of the equation as a squared term.

( x + B / 2 )^2 = C * -1 + pow( B / 2, 2 )

Take the square root of both sides.

x + B / 2 = \pmsqrt( C * -1 + pow( B / 2, 2 ) )

Isolate x to find the solution(s).

x = -B / 2\pmsqrt( C * -1 + pow( B / 2, 2 ) )

So the solutions are: x = -B / 2 + sqrt( C * -1 + pow( B / 2, 2 ) ) \text{ OR } x = -B / 2 - sqrt( C * -1 + pow( B / 2, 2 ) )

The solution is: x = -B / 2 + sqrt( C * -1 + pow( B / 2, 2 ) )

We already found the completed square: ( x + B / 2 )^2 = C * -1 + pow( B / 2, 2 )

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