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Mathematics 10 Online
OpenStudy (anonymous):

factor 21x^2-49x-9x-21

OpenStudy (anonymous):

idk how i got 7x(3x-7) 3(3x-7) and i know its wrong

hartnn (hartnn):

is the original problem 21x^2-58x-21 ?

OpenStudy (anonymous):

no

OpenStudy (anonymous):

hartnn (hartnn):

right,so thats 21x^2-40x-21=0 right ?

OpenStudy (anonymous):

yeah and then the product of 21 and -21 is -441 and then 40 so the two numbers are -49 and 9

hartnn (hartnn):

that is correct, see above u have written -49x - 9x instead of what u got now as -49x+9x

OpenStudy (anonymous):

oooh i see wow

hartnn (hartnn):

now whats common from first 2 terms in 21x^2-49x-9x-21 =0?

OpenStudy (anonymous):

7x

OpenStudy (anonymous):

i got the answer :P -3/7 and 7/3

hartnn (hartnn):

yup, correct.

OpenStudy (anonymous):

im having problems with another equation

hartnn (hartnn):

which ?

OpenStudy (anonymous):

OpenStudy (anonymous):

i got to the point 9 + or - sqrt of 1 all over 4

hartnn (hartnn):

9 from where ?

hartnn (hartnn):

Compare your quadratic equation with \(ax^2+bx+c=0\) find a,b,c then the two rots of x are: \(\huge{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\)

OpenStudy (anonymous):

yeah b is 9

hartnn (hartnn):

nopes, b is -4

OpenStudy (anonymous):

oh wait nooo i gave you the wrong equation! no wonder

hartnn (hartnn):

lol :P

OpenStudy (anonymous):

hartnn (hartnn):

still b is -9

OpenStudy (anonymous):

but in that equation it says -b so -(-9) is 9

OpenStudy (anonymous):

inside i did put put -9^2 = 81

hartnn (hartnn):

yes, u are correct \(\huge\frac{9 \pm \sqrt{1}}{4}=\frac{10}{4},\frac{8}{4}\)

OpenStudy (anonymous):

wait you are just adding and subtracting the 1?

OpenStudy (anonymous):

obviously lol never mind i feel stupid

hartnn (hartnn):

got it now ?

OpenStudy (anonymous):

yes, it was a stupid moment :P

OpenStudy (anonymous):

ok i did another one and i got 4+-sqrt of 28 all over 2 @hartnn

hartnn (hartnn):

question ?

OpenStudy (anonymous):

its the first one that sent you

OpenStudy (anonymous):

hartnn (hartnn):

thats correct :)

OpenStudy (anonymous):

but how do i factor it?

hartnn (hartnn):

do u want to factor or do u want to use quadratic formula? anyways its not factorable....

OpenStudy (anonymous):

so what would be the answer?

OpenStudy (anonymous):

i cant simplify it?

hartnn (hartnn):

all u can do is \(2 \pm \sqrt7\)

hartnn (hartnn):

\(\sqrt{28}=2\sqrt7\)

OpenStudy (anonymous):

how does that equal to that

hartnn (hartnn):

\(\sqrt{28}=\sqrt{7*4}=\sqrt{7}\sqrt{4}=2\sqrt{7}\)

OpenStudy (anonymous):

got it

OpenStudy (anonymous):

you are the best!

hartnn (hartnn):

lol. glad to hear that :) u too are awesome ;)

OpenStudy (anonymous):

lol thank you, well i gotta keep doing this stupid hw

hartnn (hartnn):

best of luck for that.

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