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Mathematics 9 Online
OpenStudy (anonymous):

quadralic formula to find any x-intercepts y=x^2-6x-6

OpenStudy (anonymous):

guadratic formula sorry mispelled

OpenStudy (anonymous):

JIthinkv

OpenStudy (anonymous):

ok... so to find the xx intercepts, you'll need to solve the equation when y=0... ie. \(\large 0=ax^2+bx+c \) can you identify the coefficients a, b, and c for me.... a= b= c=

OpenStudy (anonymous):

a=1 b=6 c=6

OpenStudy (anonymous):

but what happens if the problem say -6x-6 would it still be 0=ax^2+bx+c

OpenStudy (anonymous):

be careful... b=-6, c=-6

OpenStudy (anonymous):

that is what I was just wondering, them negative get me all of the time

OpenStudy (anonymous):

so now that you have the correct values for a, b, and c, plug them into the quadratic formula... \(\huge x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \)

OpenStudy (anonymous):

so then it would be 0=1x^2+-6x+-6 right

OpenStudy (anonymous):

yes...

OpenStudy (anonymous):

\(\huge x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-(-6)\pm \sqrt{(-6)^2-4(1)(-6)}}{2(1)} \) simplify...

OpenStudy (anonymous):

that is what I was doing wow you are fast

OpenStudy (anonymous):

ok so would you just divide all of this by 2?

OpenStudy (anonymous):

yes... the denominator is 2 hopefully some things can cancel (simplify)... let's see...

OpenStudy (anonymous):

\[3\pm \sqrt{3^{2}-(2)(12)}\]

OpenStudy (anonymous):

\(\large x=\frac{-(-6)\pm \sqrt{(-6)^2-4(1)(-6)}}{2(1)}=\frac{-6\pm \sqrt{36+24}}{2}=\frac{-6\pm \sqrt{60}}{2} \)

OpenStudy (anonymous):

you can't cancel out the 2 in the denominator after you simplify that radical first....

OpenStudy (anonymous):

wow I was way off so you would not divded the 6/2 you would just fdo the qdrt which is 36

OpenStudy (anonymous):

nooooo..... simplify the radical first....

OpenStudy (anonymous):

so what you have at the end is the answer? \[\frac{ -6\pm \sqrt{60} }{ 2 }\]

OpenStudy (anonymous):

no... that can be simplified...

OpenStudy (anonymous):

I thought that you would take it to the lowest which it would be \[\frac{ -3\pm \sqrt{30} }{2 }\]

OpenStudy (anonymous):

right

OpenStudy (anonymous):

\(\large \frac{6 \pm \sqrt{60}}{2}= \frac{6 \pm 2\sqrt{15}}{2}=\frac{\cancel{6}^3 \pm \cancel{2}^1\sqrt{15}}{\cancel{2}^1}=3\pm \sqrt{15}\)

OpenStudy (anonymous):

then I would do 30/2 right

OpenStudy (anonymous):

i think that's it....

OpenStudy (anonymous):

30/2 ????

OpenStudy (anonymous):

15

OpenStudy (anonymous):

why are you calculating 30/2 ???

OpenStudy (anonymous):

do not know but I get what you did now thanks for your help I would have never got that

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