|
|
|
= + ÷ ± ∆ ∞ × π Σ f(x) % ≤ ! ≠ * |
A perfect square trinomial is of the form
or
Here are two examples 
(note: whether you see y= or f(x) notation, the algebra is the same.)
Recall that to factor trinomials we looked for two numbers whose product was the last term and whose sum was the coefficient of the middle term. For the first equation above 3 * 3 = 9 and 3 + 3 = 6 and for the second equation -2 * -2 = 4 and -2 + -2 = -4.
Now we can see also that for a perfect square trinomial the third term is half the
middle term (coefficient) squared. 
For the first equation, half of six is 3 and 3 squared is 9 and for the second equation half of -4 is 2 and -2 squared is 4.
This idea gives us a way to solve for the zeros of quadratics that do not factor with answers that are integers. We can use the perfect square form to force the process for all quadratic equations.
REMEMBER
We want to be able to factor quadratics so that we can find the zeros of the function. We do that by setting the function equal to 0, factoring and then using the zero product rule.
Lets demonstrate with one we do know the answer to already.
Take the quadratic equation
.
[We could factor it
so we
know the zeros are -5 and -3. ]
We know that half of 8 is 4 and 4 squared is 16. But, we dont have 16 for the
third term, we have 15. So we know it is not the perfect square
.
Our procedure will still be to set the equation equal to zero and to use the zero product rule. This time, however, we will make the left side of the equation a perfect square.
![]()
We will add 1 to each side of the equation ![]()
And simplify ![]()
Now we factor ![]()
Take the square root of both sides ![]()
We must account for both roots! ![]()
So our zeros are
and ![]()
Now lets do one we cant factor.
![]()
We will add 6 to each side of the equation ![]()
And simplify ![]()
Now we factor ![]()
Take the square root of both sides ![]()
We must account for both roots! ![]()
So our zeros are
and ![]()