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y = -2x^2 - 20x - 48
then i rewrote that into: y = -2(x^2 - 20x ____) -48 -20/2 = -10 then square it and I got 100 y = -2(x^2 - 20x +100) -48
then I added 100 to both sides and I got: y + 100= -2(x - 10)^2 -48 -100 -100 ------------------------ y = -2(x - 10)^2 - 148
You subtracted 100 though ^
you add 48 to the other side and then also the 100 and the quadratic = 148
then you can factor the left side
can you show me that in steps? like from the beginning
you did it rite just set the equation = 0 then subtract 49 from both sides so 2x^2-20x=48
then complete the square and add the 100 to both sides
so it'd be 0 = -2(x^2 - 20x + 100) -48 ?
why do I need to factor that
Ok I got it thanks! and one more question
How would I solve: x^2 - 38x + c also by completing the square
c is a constant most likely just move it to the other side and complete the square
what is that going to give me?
and what would be the next step
0 = (x2 - 38x ___) +c ?
square the -38 and do the same to both sides
-38/2 = -19^2 = 361
so that is on both sides
okay so far i have: (x^2 - 38x + 361) = 361 -c
right now you should be able to factor the quadratic which is the point of completing the square
so like: ____ x___ = 361 ____+___= -38
right it will be the same number it is squared
(x-19)^2 = 361 - c?
and the first equation was x^2-10+25=49
what do I do now?
with the second you can't do anything it is most likely to show how you did the work
so just that ^
how would i write tht in intercept form
do you mean y=mx+b
no I also have to rewrite tht in intercept
the x intercept is +or-(7/5,0)
that is the y is 0