easy algebra problem i can't figure out!
Started by msu08
You need to use the quadratic equation here to find Y (remember when we have y = ax^2 + bx + c, we use the quadratic equation to find x? now, you have to use that equation to Y, knowing that a = 1 (The quoefficient of Y^2, b = 0 (because there is no y in the equation), c= -1/y^2 + 1 (the rest).
Subsitute these in [-b + (b^2 - 4ac)^(1/2)] / 2a , and I'm sure you will get the correct answer (I just don't feel like doing it here, cos it's a pain writing it all on the keyboard).
nah you don't need any quadratic. by the way can you make suer answer is not square root of x+1/-x+1
what do u do if u don't need the quadratic equation??? crack dat math gives this as the explanation which i can't follow:
x(y^2 +1)= y^2 -1
y^2 = (-x-1)/x-1
y= sqrt (x+1/x-1) this is the answer
anyone understand what they did here or another way to do this?? i cannot follow this explanation nor generate these steps on my own... help?
x(y^2 +1)= y^2 -1
y^2 = (-x-1)/x-1
y= sqrt (x+1/x-1) this is the answer
anyone understand what they did here or another way to do this?? i cannot follow this explanation nor generate these steps on my own... help?
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Could you put parenthesis in proper places so I can understand what you are asking?
2-1 is 1 and 2+1 is 3 if both numbers our powers.
is it x = (y^2) - [1/(y^2)] +1
or
x = [(y^2)-1] / [(y^2)+1]
edit: after reading the solution u posted, it seem it's the 2nd one.
Ok, this is an easy problem
when u see denominator, get rid of it first by multiplying it on both sides.
so u have x[(y^2)+1] = [(y^2)-1]
and that's same as (xy^2) + x = (y^2) -1
and that's same as (xy^2) + x - (y^2) +1 = 0
and that's same as (xy^2)-(y^2)+x+1 = 0
and that's same as y^2(x-1) +x+1 =0
and that's same as y^2 (x-1) = -x-1
and that's same as y^2 = (-x-1)/(x-1)
and that's same as y = square root of (-x-1)/(x-1)
that's the best I can explain.
It's been a long time but I think i learned about 6 yrs ago that (-x-1) is same as (x+1) but I don't know how that works now...unless (-x-1) = -(x+1) but this one doesn't even have negative sign in front.
can someone explain this very last step why (-x-1) can become (x+1)?
or
x = [(y^2)-1] / [(y^2)+1]
edit: after reading the solution u posted, it seem it's the 2nd one.
Ok, this is an easy problem
when u see denominator, get rid of it first by multiplying it on both sides.
so u have x[(y^2)+1] = [(y^2)-1]
and that's same as (xy^2) + x = (y^2) -1
and that's same as (xy^2) + x - (y^2) +1 = 0
and that's same as (xy^2)-(y^2)+x+1 = 0
and that's same as y^2(x-1) +x+1 =0
and that's same as y^2 (x-1) = -x-1
and that's same as y^2 = (-x-1)/(x-1)
and that's same as y = square root of (-x-1)/(x-1)
that's the best I can explain.
It's been a long time but I think i learned about 6 yrs ago that (-x-1) is same as (x+1) but I don't know how that works now...unless (-x-1) = -(x+1) but this one doesn't even have negative sign in front.
can someone explain this very last step why (-x-1) can become (x+1)?
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Should be sqrt of (x+1)/(1-x) or sqrt of (-x-1) / (x-1). They are the same.
But sqrt of (x+1) / (x-1) is not correct.
-x-1 becomes x+1 when you multiply it by -1.
But sqrt of (x+1) / (x-1) is not correct.
-x-1 becomes x+1 when you multiply it by -1.
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