A handy formula easily memorized by repetition. It is used in many instances, such as trying to find the roots of an equation, or solving an equation that will not factor with integers.
To solve using the Quadratic Formula:
In any equation
ax^2+bx+c,
The roots can be found by substituting into:
-b(+,-)sqrt(b^2-4ac)/2a
OR
Negative "B" plus or minus the square root of "B" squared minus 4(a)(c), all divided by 2(a).
To solve using the Quadratic Formula:
In any equation
ax^2+bx+c,
The roots can be found by substituting into:
-b(+,-)sqrt(b^2-4ac)/2a
OR
Negative "B" plus or minus the square root of "B" squared minus 4(a)(c), all divided by 2(a).
To solve using the Quadratic Formula:
In any equation
ax^2+bx+c,
The roots can be found by substituting into:
-b(+,-)sqrt(b^2-4ac)/2a
OR
Negative "B" plus or minus the square root of "B" squared minus 4(a)(c), all divided by 2(a).
EX:
Hm..
10x^2+13X-5
-13(+,-)sqrt(169+200)/20
>_>
In other words, by substituting the numbers in, you get 13(+,-)sqrt(369)/20.
As you may or may not have guessed, this gives you a very nasty number, which should be approximated and rounded to the nearest thousandth.
In any equation
ax^2+bx+c,
The roots can be found by substituting into:
-b(+,-)sqrt(b^2-4ac)/2a
OR
Negative "B" plus or minus the square root of "B" squared minus 4(a)(c), all divided by 2(a).
EX:
Hm..
10x^2+13X-5
-13(+,-)sqrt(169+200)/20
>_>
In other words, by substituting the numbers in, you get 13(+,-)sqrt(369)/20.
As you may or may not have guessed, this gives you a very nasty number, which should be approximated and rounded to the nearest thousandth.
by "Seagaia" May 21, 2005
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