Ask your own question, for FREE!
Mathematics 21 Online
OpenStudy (anonymous):

The quadratic x2 + 5x – 14 has what type of roots? Two Irrational Two Complex Two Rational One Rational Double

OpenStudy (anonymous):

factor that bad boy

OpenStudy (anonymous):

idk how

OpenStudy (anonymous):

two numbers that multiply to give you -14 and add to give you 5

OpenStudy (anonymous):

Iwhat are they

OpenStudy (anonymous):

?

myininaya (myininaya):

\[b^2-4ac=?\]

OpenStudy (anonymous):

yeah i got -8

OpenStudy (anonymous):

In general ax^2+bx+c you can factor by getting the two numbers that multiply to give you c and add to give you b

OpenStudy (anonymous):

In this case -2 and +7

OpenStudy (anonymous):

So it factors like (x-2)(x+7)

OpenStudy (anonymous):

therefore roots are 2 and -7

myininaya (myininaya):

\[b^2-4ac \text{ is a perfect square but not 0 then both solutions are rational}\] \[b^2-4ac \text{ is negative then both solutions are imaginary }\] \[b^2-4ac \text{ is not a perfect square then both solutions are irrational} \] \[b^2-4ac \text{ is 0 then there is one solution with multiplicity 2 }\]

OpenStudy (anonymous):

i just need to find out what type of root -8 is

OpenStudy (anonymous):

Its better to factor and see for himself

OpenStudy (anonymous):

thery are both rational b^2-4ac does not equal -8

myininaya (myininaya):

@prestonchatman37 so you got \[b^2-4ac \text{ is negative } ?\] Then based on what I said above what can you say about your solutions?

OpenStudy (anonymous):

5^2-4(1)(-14)=81

myininaya (myininaya):

Yes that looks better what @Bitmaximus has So \[b^2-4ac =81 \] is 81 a perfect square or not?

OpenStudy (anonymous):

it is

myininaya (myininaya):

I wanted him to answer but ok lol

OpenStudy (anonymous):

Haha lol and I just wanted him to factor the polynomial

OpenStudy (anonymous):

:P

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!