First comes the quadratic equation, then comes the quadratic formula. The quadratic formula is
the solution to the quadratic equation:
2
a x + bx + c = 0
(1-8)
in which:
x is the variable whose value is sought, and
a, b, and c are constants
The goal is to find the value of x that makes the left side 0. That value is given by the quadratic
formula:
− b ± b2 − 4ac
x =
(1-9)
a
2
to be read/said:
‘x’ equals minus ‘b’, plus-or-minus the square root of ‘b’ squared
minus four ‘a’ ‘c’, all over two ‘a’.
So, how do you know when you have to use the quadratic formula? There is a good chance that
you need it when the square of the variable for which you are solving, appears in the equation
you are solving. When that is the case, carry out the algebraic steps needed to arrange the terms
as they are arranged in equation 1-8 above. If this is impossible, then the quadratic formula is
not to be used. Note that in the quadratic equation you have a term with the variable to the
second power, a term with the variable to the first power, and a term with the variable to the
zeroth power (the constant term). If additional powers also appear, such as the one-half power
(the square root), or the third power, then the quadratic formula does not apply. If the equation
includes additional terms in which the variable whose value is sought appears as the argument of
a special function such as the sine function or the exponential function, then the quadratic
formula does not apply. Now suppose that there is a square term and you can get the equation
that you are solving in the form of equation 1-8 above but that either b or c is zero. In such a
case, you can use the quadratic formula, but it is overkill. If b in equation 1-8 above is zero then
the equation reduces to
2
a x + bx = 0
The easy way to solve this problem is to recognize that there is at least one x in each term, and to
factor the x out. This yields:
(a x + b)x = 0
Then you have to realize that a product of two multiplicands is equal to zero if either
multiplicand is equal to zero. Thus, setting either multiplicand equal to zero and solving for x
yields a solution. We have two multiplicands involving x, so, there are two solutions to the
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