Solve: X^4-13x^2+36=0.
I can't figure this out.
Also,
Y^(2/3)-2y^(1/3)-8=0
for the second one, can I add 8 to the right side, cube both sides and end up with
Y^2-2y=512, then use the quadratic formula?
Solve: X^4-13x^2+36=0.
I can't figure this out.
Also,
Y^(2/3)-2y^(1/3)-8=0
for the second one, can I add 8 to the right side, cube both sides and end up with
Y^2-2y=512, then use the quadratic formula?
I tried two ways now and for some reason am still having a hard time solving.
The first way, was to factor X^2 from the first term, (a), and X from the second, (b). And so I have:
X^2(X^2)-13x(x)+36=0 A=x^2 B=13x, and C=36
The second way was to factor change the first term, x^4, into X^3(x). And so I have:
-13x^2+X^3(x)+36=0 A=-13 B=x^3, and C=36
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When I plug these values into the quadratic equation:A=x^2, B=13x, C=36
x=-13+or-[(169-144x^2)/2x^2]^-2 I solve as:
X=-13+or-13-12(2^-2) X=-13+or-(-3.970...)
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When I use the other values: A=-13 B=X^3 and C=36
x=-x^3+or-[(X^6+1872)/-26]^-2 I have a negative inside a square root.
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Am I doing this wrong? Or am I doing it right, but making algebraic errors when simplifying?
$\displaystyle X^4-13x^2+36=0$
$\displaystyle (X^2)^2 - 13X^2 + 36 = 0 $
{instead "x" you have square of it, take it easy}
$\displaystyle (X^2 - 9) (X^2 - 4) = 0$
$\displaystyle X^2 - 9 = 0$ OR $\displaystyle X^2 - 4 = 0$
$\displaystyle X^2 = 9$ OR $\displaystyle X^2 = 4$
$\displaystyle X = 3, -3, 2, -2$
Hello StudentMCCS,
Looking at your post in which you were having trouble solving quadratics it is best to use the factoring method before going to the quadratic formula. In these two problems each can be solved this way.You should learn it if not already known.