# Khan/khan-exercises

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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.

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