#Thinking
21 messages · Page 1 of 1 (latest)
I was just thinking, wouldn't having a fraction as a b or c make it unfactorable and you'd have to use the quadratic formula? Where x never = a whole number/
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he's shown u an example of when the coefficients are a fraction but the roots are not, try using the quadratic formula on 0.5x^2-4x+0 and see what happens. You can have fractional or integer roots with either integer or fractional coefficients
I can show you another way to see this problem.
There are these things called Viéte-formulas
they say that in a quadratic ax^2+bx+c=0 there are two equations always ture
-b/a = x_1 + x_2 and c/a = x_1*x_2
Consider these two equations and you can see that there are infinitely many cases where a can be a fraction and x_1 and x_2 are integers
On the factorisation
you have to pull out the coefficient of the highest order term and proceed after that
eg
0.5x^2+4x = 0.5(x^2 + 8x)
Now you can factorize and you are going to have a 0.5 term at the beginning