# public Khan /khan-exercises

### Subversion checkout URL

You can clone with HTTPS or Subversion.

Fetching contributors…

Cannot retrieve contributors at this time

file 141 lines (131 sloc) 6.954 kb
 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140         Solving quadratics by factoring 2

randRangeNonZero( -10, 10 )            randRangeNonZero( -10, 10 )
randRange(1,6)

SQUARE*A*B                A*B                SQUARE*(-A-B)                -A-B

Determine where f(x) intersects the x-axis.

f(x) = plus(SQUARE + "x^2") + plus( LINEAR + "x" ) + CONSTANT

A

B

x = {}\space \text{and} \space x = {}

The two numbers -A and -B satisfy both conditions:

\qquad \color{PINK}{-A} + \color{PINK}{-B} =                        \color{GREEN}{SIMPLELINEAR}

\qquad \color{PINK}{-A} \times \color{PINK}{-B} =                        \color{BLUE}{SIMPLECONSTANT}

So (x A < 0 ? "+" : "" \color{PINK}{-A})                    (x B < 0 ? "+" : "" \color{PINK}{-B}) = 0.

Since (x A < 0 ? "+" : "" -A)                    (x B < 0 ? "+" : "" -B) = 0,                    we know that one or both quantities must equal zero for the equation to be true.

x + -A = 0 or x + -B = 0

Thus, x = A and x = B are the solutions.

SQUARE * A * A                A * A                SQUARE * ( -2 * A )                -2 * A

Determine where f(x) intersects the x-axis.

f(x) = plus( SQUARE + "x^2") + plus( LINEAR + "x" ) + CONSTANT

The number -A used twice satisfies both conditions:

\qquad \color{PINK}{-A} + \color{PINK}{-A} =                        \color{GREEN}{SIMPLELINEAR}

\qquad \color{PINK}{-A} \times \color{PINK}{-A} =                        \color{BLUE}{SIMPLECONSTANT}

So (x A < 0 ? "+" : "" \color{PINK}{-A})^2 = 0.

x + -A = 0

Thus, x = A is the solution.

The function intersects the x-axis where f(x) = 0, so solve the equation:

\qquad                plus( SQUARE + "x^2" )                LINEAR >= 0 ? "+" : ""                plus( "\\color{" + GREEN + "}{" + LINEAR + "}x" )                CONSTANT >= 0 ? "+" : ""                plus( "\\color{" + BLUE + "}{" + CONSTANT + "}" )                = 0

The function intersects the x-axis where f(x) = 0, so solve the equation:

\qquad plus( SQUARE + "x^2", LINEAR + "x", CONSTANT ) = 0

Dividing both sides by SQUARE gives:

\qquad x^2                SIMPLELINEAR >= 0 ? "+" : ""                plus( "\\color{" + GREEN + "}{" + SIMPLELINEAR + "}x" )                SIMPLECONSTANT >= 0 ? "+" : ""                plus( "\\color{" + BLUE + "}{" + SIMPLECONSTANT + "}" )                = 0

The coefficient on the x term is SIMPLELINEAR            and the constant term is SIMPLECONSTANT, so we need to find two numbers            that add up to SIMPLELINEAR and multiply to            SIMPLECONSTANT.

Something went wrong with that request. Please try again.