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The solutions for some quadratic equations are not rational, and cannot be obtained by factoring.
The quadratic formula, however, may be used to solve ANY quadratic equation (even the ones that can be factored). This is a formula that you want to know and remember!

reminder
• As you can see in the formula, the coefficients (numbers) "a", "b", and "c" from
    ax2 + bx + c = 0 are substituted into the formula.
• Also notice that the formula will yield two (±) solutions, since a quadratic is a
    second degree equation.
• The "2a" in the denominator is underneath the entire top, not just the radical.
• Be careful with the "signs" of the "a", "b", and "c" values when substituting.

Let's see the Quadratic Formula at work in various situations:

expin1    Solve: x2 + 2x - 15 = 0
Also factorable.
Solution using Quadratic Formula: a = 1; b = 2; c = -15
quadform1
This equation is also factorable.
(x + 5)(x - 3) = 0
x = -5; x = 3

Notice that the quadratic formula ALSO gives the correct results.

beware1
Notice the parentheses around
the -15 to avoid confusion.



expin1   Solve: 2x2 - 10x = -3
Not set equal to zero!
Set equation equal to zero: 2x2 -10x + 3 = 0
Solution using Quadratic Formula: a = 2; b = -10; c = 3
quadform2a
quadform22b
Not factorable.

Notice the needed parentheses for dealing with the "b" value of "-10".

quadbeware2a
Radical answers are"exact" answers.
Decimal answers are
"approximate" answers most often used in applied word problems.



expin1   Solve: x2 -10x + 25 = 0
Repeated answer!
Solution using Quadratic Formula: a = 1; b = -10; c = 25

quadform4
Be careful here! It appears that there is only ONE answer, but there is actually a "repeated" answer.

This equation is also factorable.
(x - 5)(x - 5) = 0
x = 5; x = 5
Answer repeats.
beware4



expin1   Solve: x2 - 6x + 13 = 0
Negative under radical!
Solution using Quadratic Formula: a = 1; b = -6; c = 13

quadform3
Not factorable.

quadbeware2

quadform3a

 

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