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Mathematics 15 Online
OpenStudy (anonymous):

how to slove 18 b?

OpenStudy (skullpatrol):

slove?

OpenStudy (anonymous):

h(x) = x2 – 5x – 11. ??

OpenStudy (skullpatrol):

$$\huge x^2 -5x-11$$

OpenStudy (anonymous):

how to solve it if h(x) is 0

OpenStudy (anonymous):

put x=0

OpenStudy (skullpatrol):

"if h(x) is 0" means h(x) = 0 "put x = 0" means h(0)

OpenStudy (unklerhaukus):

h(x) = x^2 – 5x – 11 = 0

OpenStudy (skullpatrol):

h(0) = 0^2 -5*0 - 11 = -11

OpenStudy (unklerhaukus):

one question at a time

OpenStudy (anonymous):

h(x)=0 it means that x=0

OpenStudy (skullpatrol):

@HARSH123 how does it mean that?

OpenStudy (skullpatrol):

When you write: "h(x)=0" No value has been given to x, but the whole equation has been set equal to 0, right?

OpenStudy (skullpatrol):

So, h(x)=0 does not mean that x=0.

OpenStudy (anonymous):

but the answer is sayin -1.65 or 6.55 o.o how is that possible?

OpenStudy (skullpatrol):

h(0) means x=0

OpenStudy (skullpatrol):

@luzala Do you know the quadratic formula?

OpenStudy (anonymous):

then what i the answer can u tell me

OpenStudy (unklerhaukus):

\[ax^2+bx+c=0\\\,\\x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

OpenStudy (anonymous):

yeahhh.. @skullpatrol we have to use that?

OpenStudy (anonymous):

ohhhh .... got it

OpenStudy (anonymous):

OMG thankyou @skullpatrol

OpenStudy (skullpatrol):

Thanks for trying to understand :)

OpenStudy (anonymous):

:D

OpenStudy (skullpatrol):

You have to use the quadratic formula because you can't factor $$ x^2 – 5x – 11$$ since 11 is a prime number with factors of 1 and 11 which don't add up to 5, right?

OpenStudy (anonymous):

yes... iam trying that out ryt now..

OpenStudy (skullpatrol):

This kind of polynomial is called a "prime polynomial," since it is an irreducible polynomial with integral coefficients whose greatest monomial factor is 1.

OpenStudy (anonymous):

ok! but i tried the quardratic formula but the ans is not comin...

OpenStudy (skullpatrol):

a = 1, b = -5, and c = -11

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