randRangeNonZero( -10, 10 ) randRangeNonZero( -10, 10 )
1
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 = {}` and `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 = \quad` 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:

`x^2 + plus(SIMPLELINEAR + "x") + 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`.