0
x10 = x5.x5 = (x5)2 [x power five whole squared]
Equation is
x10 + x5 - 2
Replacing x10
(x5)2+x5 - 2
Substituting x5 by Y
Equation becomes
Y2 + Y - 2 = 0
Y2 + 2Y - Y - 2 = 0
Y(Y + 2) - 1(Y+2) = 0
(Y-1)(Y+2) = 0
The values of Y are 1 and -2
Y = x5 = 1
Therefore, x = 1
Y = x5 = -2
x = fifth root of -2, which is an imaginary value.
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Why you solve a differential equation for x?
In its normal form, you do not solve differential equation for x, but for a function of x, usually denoted by y = f(x).
How do you solve x plus 1 over x?
You don't "solve" an expression. You can solve an equation; an expression can be simplified or otherwise manipulated, and if you know the value of "x" (in this case), you can evaluate its value.
What variable is used for y intercept?
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