randRangeNonZero(-10, 10) randRangeNonZero(-10, 10)
randRange(1, 6)
SQUARE * A * B A * B SQUARE * (-A - B) -A - B

Solve for `x`:

`plus(SQUARE + "x^2") + plus(LINEAR + "x") + CONSTANT = 0`

A
B
`x = {}` or `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}`

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

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

`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

Solve for `x`:

`plus( SQUARE + "x^2") + plus( LINEAR + "x" ) + CONSTANT = 0`

`x = {}`A

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 + \color{PINK}{-A})^2 = 0`.

So `(x - \color{PINK}{A})^2 = 0`.

`x + -A = 0`

Thus, `x = A` is the solution.

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`.