quadralic formula to find any x-intercepts y=x^2-6x-6
guadratic formula sorry mispelled
JIthinkv
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=
a=1 b=6 c=6
but what happens if the problem say -6x-6 would it still be 0=ax^2+bx+c
be careful... b=-6, c=-6
that is what I was just wondering, them negative get me all of the time
so now that you have the correct values for a, b, and c, plug them into the quadratic formula... \(\huge x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \)
so then it would be 0=1x^2+-6x+-6 right
yes...
\(\huge x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-(-6)\pm \sqrt{(-6)^2-4(1)(-6)}}{2(1)} \) simplify...
that is what I was doing wow you are fast
ok so would you just divide all of this by 2?
yes... the denominator is 2 hopefully some things can cancel (simplify)... let's see...
\[3\pm \sqrt{3^{2}-(2)(12)}\]
\(\large x=\frac{-(-6)\pm \sqrt{(-6)^2-4(1)(-6)}}{2(1)}=\frac{-6\pm \sqrt{36+24}}{2}=\frac{-6\pm \sqrt{60}}{2} \)
you can't cancel out the 2 in the denominator after you simplify that radical first....
wow I was way off so you would not divded the 6/2 you would just fdo the qdrt which is 36
nooooo..... simplify the radical first....
so what you have at the end is the answer? \[\frac{ -6\pm \sqrt{60} }{ 2 }\]
no... that can be simplified...
I thought that you would take it to the lowest which it would be \[\frac{ -3\pm \sqrt{30} }{2 }\]
right
\(\large \frac{6 \pm \sqrt{60}}{2}= \frac{6 \pm 2\sqrt{15}}{2}=\frac{\cancel{6}^3 \pm \cancel{2}^1\sqrt{15}}{\cancel{2}^1}=3\pm \sqrt{15}\)
then I would do 30/2 right
i think that's it....
30/2 ????
15
why are you calculating 30/2 ???
do not know but I get what you did now thanks for your help I would have never got that
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