Express p in terms of q if [tex]p = {x}^{2} + \frac{1}{

Ответ:

nuna37

11.03.2022

$p=x^2+\dfrac{1}{x^2}\\\\p=\dfrac{x^4}{x^2}+\dfrac{1}{x^2}\\\\p=\dfrac{x^4+1}{x^2}\qquad(*)$

$q=x+\dfrac{1}{x}\\\\q=\dfrac{x^2}{x}+\dfrac{1}{x}\\\\q=\dfrac{x^2+1}{x}\qquad\text{square both sides}\\\\q^2=\left(\dfrac{x^2+1}{x}\right)^2\\\\q^2=\dfrac{(x^2+1)^2}{x^2}\qquad\text{use}\ \ (a+b)^2=a^2+2ab+b^2\\\\q^2=\dfrac{(x^2)^2+2(x^2)(1)+1^2}{x^2}\\\\q^2=\dfrac{x^4+2x^2+1}{x^2}\\\\q^2=\dfrac{x^4+1}{x^2}+\dfrac{2x^2}{x^2}\\\\q^2=\dfrac{x^4+1}{x^2}+2\qquad\text{subtract 2 from both sides}\\\\q^2-2=\dfrac{x^4+1}{x^2}\\\\\text{From (*) we have}\\\\\boxed{p=q^2-2}$

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Ответ:

christson805

10.09.2022

Mathematics

(m + 5)(m - 9) = 0

Step-by-step explanation:

By setting them to zero, you will get the following solutions.

m + 5 = 0 m - 9 = 0

m = -5 m = 9

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Express p in terms of q if $p = {x}^{2} + \frac{1}{ {x}^{2} } \: and \: \: q = x + \frac{1}{x}$

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