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OpenStudy (anonymous):

PRE-CALC HELP: What is the value of b? 5x2 - 3y2 - 20x + 6y + 2 = 0 a. √5 b. 5 c. 15 d. -15

OpenStudy (johnweldon1993):

\[\large 5x^2 - 3y^2 - 20x + 6y + 2 = 0\] value of 'b' ?

OpenStudy (anonymous):

@johnweldon1993 it won't let me message you because you aren't a fan of mine..:( that's why I didn't message back. And yeah its confusing, pre-calc

OpenStudy (anonymous):

What is the value of c? 5x2 - 3y2 - 20x + 6y + 2 = 0 Select one: a. 2 b. 4 c. √4 d. √8 This was an older question and someone said the answer was D and I got it right.

OpenStudy (johnweldon1993):

O.o my settings have never been like that O.o odd...okay hang on 1 second and let me work with that and I'll be right back for this problem :D

OpenStudy (anonymous):

Okay! Yeah it's really weird I don't get it :// I was like aw

OpenStudy (johnweldon1993):

So we have \[\large 5x^2 - 3y^2 - 20x + 6y + 2 = 0\] subtract 2 from both sides of the equation \[\large 5x^2 - 3y^2 - 20x + 6y = -2\] Group together the 'x' terms and the 'y' terms \[\large (5x^2 - 20x) - (3y^2 + 6y) = -2\] Factor out what is common between the parenthesis...for the first set (5x^2 - 20x) we can factor out a 5 and the second set factor out a 3 \[\large 5(x^2 - 4x) - 3(y^2 + 2y) = -2\] We'll want to divide everything by 5 now \[\large (x^2 - 4x) - \frac{3}{5}(y^2 + 2y) = \frac{-2}{5}\] And then divide everything by 3 as well \[\large \frac{1}{3}(x^2 - 4x) - \frac{1}{5}(y^2 + 2y) = -\frac{2}{15}\] Now we need to complete some squares here \[\large \frac{1}{3}(x^2 - 4x + 4) - \frac{1}{5}(y^2 + 2y + 1) = -\frac{2}{15} + \frac{1}{9}(4) - \frac{1}{5}(1)\] \[\large \frac{1}{3}(x^2 - 4x + 4) - \frac{1}{5}(y^2 + 2y + 1) = \frac{1}{9}\] So altogether we have \[\large \frac{(x - 2)^2}{3} - \frac{(y + 1)^2}{5} = \frac{1}{9}\] Now we want that 1/9 to become 1...so we need to multiply everything by 9 \[\large \frac{9(x - 2)^2}{3} - \frac{9(y + 1)^2}{\color \red{5}} = 1\] and now lets compare that to the general form \[\large \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{\color \red{b^2}} = 1\] Looky there...now we know that \(\large b^2 = 5\) or \(\large b = \sqrt{5}\)

OpenStudy (anonymous):

WELL HOLY CRAP D;

OpenStudy (johnweldon1993):

Lol I know! ;P

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