Determine where f(x)
intersects the x
-axis.
f(x) = plus(SQUARE + "x^2") + plus( LINEAR + "x" ) + CONSTANT
x=A
and x=B
x=-A
and x=-B
x=B
and x=A+B
x=A
onlyx=B
onlyx=SQUARE
and x=LINEAR
The two numbers -A
and -B
satisfy both conditions:
-A + -B = SIMPLELINEAR
-A \cdot -B = SIMPLECONSTANT
So (x + -A)(x + -B) = 0
.
x + -A = 0
or x + -B = 0
Thus, x = A
and x = B
are the solutions.
Determine where f(x)
intersects the x-axis.
f(x) = plus( SQUARE + "x^2") + plus( LINEAR + "x" ) + CONSTANT
x=A
only
x=-A
and x=A
x=2 * A
and x=-A
x=-2 * A
and x=A
x=-A
onlyThe number -A
used twice satisfies both conditions:
-A + -A = SIMPLELINEAR
-A \cdot -A = SIMPLECONSTANT
So (x + -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:
plus( SQUARE + "x^2" ) + plus( LINEAR + "x") + CONSTANT=0
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
.