3x^2-2x+7= 0 how do we factor this?
Find the roots of the equation. The factors will be (x-root one)(x-root two)
yea it doesn't work that way. I tried factoring by grouping mertsj
This doesn't factorise.
it does , it will work using comp[leteing the square method. but I need a simpler way
Maybe I misunderstand what you are trying to achieve here. Do you want to express this as (something)(something)=0?
if the discriminant value (D) < 0, obviously it cant be factored. u need the quadratic formula to solve this. x = (-b +- sqrt(D))/2a with D = discriminant = b^2 - 4ac
@Bad2zBone yup ! or solve for x
This equation can not be solved for real x.
it cn because regardless lets say it was (x-3) (x+2) x will equal +3 and -2
You can only factorise if you have real roots. In this case there is no x value that will satisfy this equation and hence you can't factorise this.
oh ok I understrand what you are saying now thanks !
completing the square I personally think is the BEST method
\[x=\frac{2\pm \sqrt{4-4(3)(7)}}{6}=\frac{2\pm \sqrt{-80}}{6}=\frac{1\pm2\sqrt{5}i}{3}\]
From 5 employees at a company, a group of 3 employees will be chosen to work on a project. How many different groups of 3 employees can be chosen? can u hlp me with this? I keep getting 5
So the factors are: \[[x-(\frac{1+2\sqrt{5}i}{3})][x-(\frac{1-2\sqrt{5}i}{3})]\]
5cr 3 is 5x4x3/(3x2x1)(2x1 right?
\[\frac{5!}{3!2!}=\frac{5\times4}{2}=10\]
Well he hasn't mentioned the use of imaginary numbers, so.. The question is not well defined.
@Mertsj can u explain why is it only 5x4 not 5x4x3
im a she lol but im sorry @Bad2zBone
Because the 3! cancels
exactly but the 2 does not on the bottom so it should be over 4 . but m probably wrong
\[\frac{5!}{3!2!}=\frac{5\times4\times3!}{3!2!}=\frac{5\times4}{2!}=\frac{20}{2\times1}=\frac{20}{2}=10\]
2! = 2 x 1
oh ok thankyou both soooo much :D
yw
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