Jarrett180
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When you are trying to solve Quadratic Equations, do you try to find factors of C that add to make B?
Quadratic Equations
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Jarrett180
New member
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- Sep 19, 2011
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Re:re:
So if they are all adding together and all of the numbers are posative, then it wont work, right?
So if they are all adding together and all of the numbers are posative, then it wont work, right?
No, that's not what I mean. If all of the numbers are positive, and the factors of c can be added together to equal b, then you have found the correct factors.
JeffM - I believe that Jarrett is referring to the fact that often, a quadratic equation can be factored easily. The factors of c will point you towards the correct factors. The combinations of b and c being positive or negative will indicate the final answer.
To use an example from the link you provided:
x^2 - 4x + 4 = 0
(x-2)(x-2) = 0
x = 2
This may not be applicable for all quadratic equations, but I often see the answer in these terms before I even think of the quadratic formula.
Note: edited to clarify the views stated in this post are my own opinion and the opinion of the voices in my head, and may not reflect reality...
Last edited:
IF you are dealing with a quadratic expression of the form
\(\displaystyle ax^2 + bx + c\)
where a = 1, THEN you look for two factors of c whose sum is b.
For example, if you are trying to factor
\(\displaystyle x^2 - 5x + 6\)
the coefficient of the squared term is 1. The two factors of +6 which add up to -5 are (-2) and (-3), so you can write the quadratic as the product of two binomials as follows:
(x - 2)(x - 3)
However!!!! This does not ALWAYS "work."
IF the coefficient of the \(\displaystyle x^2\) term is something OTHER THAN 1, the process becomes a bit more difficult (and this is the cause of many errors and much confusion when people try to apply the "old" process to a "new form" of the equation).
Here's an example of what I mean.
\(\displaystyle 6x^2 - 7x - 5\)
If you ignore the coefficient of x2, and look only at "b" and "c" in this quadratic trinomial, you'd be trying to find two factors of -5 which add up to -7. And obviously, there aren't any integers which are going to do the job. So one might assume that this quadratic can't be factored using integers only. One would, in fact, be WRONG in that assumption.
Here's an approach which always (ALWAYS) works for any quadratic trinomial that's factorable over the integers.
1) Multiply "a" and "c". In the above example, that means "multiply 6 and (-5)," which gives -30.
2) Look for factors of this product (-30 in our example) whose sum is "b" (the middle coefficient, -7 in our example). Since (-10 )* 3 = -30, and (-10) + 3 = -7, the factors we want are -10 and +3.
3) Rewrite the middle term of the quadratic trinomial, using the factors found in step 2 to replace the middle coefficient:
\(\displaystyle 6x^2 - 10x + 3x - 5\)
4) Now, you can group the first two terms together, and the last two terms together, and remove a common factor from each group:
2x(3x - 5) + 1(3x - 5)
(3x - 5)(2x + 1)
The factoring is complete.
The "check" is to multiply the two factors together; if this results in The "check" is to multiply the two factors together; if this results in The "check" is to multiply the two factors together; if this results in The "check" is to multiply the two factors together; if this results in the original quadratic trinomial, you know you've done it correctly.
\(\displaystyle ax^2 + bx + c\)
where a = 1, THEN you look for two factors of c whose sum is b.
For example, if you are trying to factor
\(\displaystyle x^2 - 5x + 6\)
the coefficient of the squared term is 1. The two factors of +6 which add up to -5 are (-2) and (-3), so you can write the quadratic as the product of two binomials as follows:
(x - 2)(x - 3)
However!!!! This does not ALWAYS "work."
IF the coefficient of the \(\displaystyle x^2\) term is something OTHER THAN 1, the process becomes a bit more difficult (and this is the cause of many errors and much confusion when people try to apply the "old" process to a "new form" of the equation).
Here's an example of what I mean.
\(\displaystyle 6x^2 - 7x - 5\)
If you ignore the coefficient of x2, and look only at "b" and "c" in this quadratic trinomial, you'd be trying to find two factors of -5 which add up to -7. And obviously, there aren't any integers which are going to do the job. So one might assume that this quadratic can't be factored using integers only. One would, in fact, be WRONG in that assumption.
Here's an approach which always (ALWAYS) works for any quadratic trinomial that's factorable over the integers.
1) Multiply "a" and "c". In the above example, that means "multiply 6 and (-5)," which gives -30.
2) Look for factors of this product (-30 in our example) whose sum is "b" (the middle coefficient, -7 in our example). Since (-10 )* 3 = -30, and (-10) + 3 = -7, the factors we want are -10 and +3.
3) Rewrite the middle term of the quadratic trinomial, using the factors found in step 2 to replace the middle coefficient:
\(\displaystyle 6x^2 - 10x + 3x - 5\)
4) Now, you can group the first two terms together, and the last two terms together, and remove a common factor from each group:
2x(3x - 5) + 1(3x - 5)
(3x - 5)(2x + 1)
The factoring is complete.
The "check" is to multiply the two factors together; if this results in The "check" is to multiply the two factors together; if this results in The "check" is to multiply the two factors together; if this results in The "check" is to multiply the two factors together; if this results in the original quadratic trinomial, you know you've done it correctly.