If 2((x2)+1)=5x, find i)x−(1x)ii)x3−(1x3)

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Corben Pittman · Accepted Answer

We need to first find the value of xx by solving $2 (x^{2} + 1) = 5 x .$
To solve $2 (x^{2} + 1) = 5 x$ , distribute the 2 to get $2 x^{2} + 2 = 5 x .$
Subtracting 5x5x on both sides then gives $2 x^{2} - 5 x + 2 = 0 .$
Factoring would then gives $(2 x - 1) (x - 2) = 0 .$
Now that the equation is in factored form, we can set each factor equal to 0 to find the possible values of x:
2x−1=0 \ x−2=0
2x=1 \ x=2
$x = \frac{1}{2}$
Now that we have the possible values of x, we can substitute them into each expression and then simplify to find the values:
$i) I f x = \frac{1}{2}$ , then $x - (\frac{1}{x}) = \frac{1}{2} - (\frac{1}{1} / 2) = \frac{1}{2} - 2 = - (\frac{3}{2})$
If x=2, then $x - (\frac{1}{2}) = 2 - (\frac{1}{2}) = \frac{3}{2}$
ii)If $x = \frac{1}{2}, t h e n (x^{3}) - (\frac{1}{x^{3}}) = {(\frac{1}{2})}^{3} - (\frac{1}{{(\frac{1}{2})}^{3}}) = \frac{1}{8} - (\frac{1}{1} / 8) = ((\frac{1}{8}) - 8) = - (\frac{63}{8})$
If x =2, then $(x^{3}) - (\frac{1}{x^{3}}) = (2^{3}) - (\frac{1}{2^{3}}) = 8 - (\frac{1}{8}) = \frac{63}{8}$

If 2((x^2)+1)=5x, find i) x−(1/x) ii) x^3−(1/x^3)

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