Special Products Question:$ (p+\frac1p)=5, \ (p-(\frac1p))^2=?$

Question

This is a special productions Question. $ (p+\frac1p)=5, \ (p-(\frac1p))^2=?$

Anyone has solution?

Please answer fast if you know. I have exam tomorrow. I tried to square both sides and use formula $a^2\ + 2ab+ b^2$ but could do only half.

Answer 1

You have

$$ \left(p + \frac{1}{p} \right)^2 = p^2 + \frac{1}{p^2} + 2 $$

and

$$ \left(p - \frac{1}{p} \right)^2 = p^2 + \frac{1}{p^2} - 2 $$

Therefore $$ \left(p - \frac{1}{p} \right)^2 = \left(p + \frac{1}{p} \right)^2 - 4 = 21 $$

Answer 2

HINT: $$p+\frac1p=5\implies p^2-5p+1=0$$ Solve using the quadratic formula and substitute into your desired expression.

Answer 3

Answer:

$$\boxed{ \ \mathbf{\left(p-\frac1p\right)^2 = 21} \ }$$

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Proof:

Write the following formula:

$$\begin{align} p^2+\left(\frac1p\right)^2 &= \left(p+\frac1p\right)^2 - 2\cdot p\cdot\left(\frac1p\right).\end{align}$$

By substituting $p +\dfrac1p = 5$, it follows that $$\begin{align}p^2 + \left(\frac1p\right)^2 &= 5^2 - 2 \\ &= 23.\end{align}$$ Therefore, $$\begin{align}\left(p-\frac1p\right)^2 &= 23 - 2\cdot p\cdot\left(\frac1p\right)\\ &= \text{what exactly?}\end{align}$$

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Reasoning:

We begin by finding the value of $p^2 + \left(\dfrac1p\right)^2$ because it is a common constant, such that

$$\left(p\pm\frac1p\right)^2 = \color{blue}{p^2 +\left(\frac1p\right)^2} \pm 2$$

Answer 4

$$(x+x^{-1})^2=x^2+2+x^{-2}$$ $$(x-x^{-1})^2=x^2-2+x^{-2}$$ Therefore $$(x-x^{-1})^2=(x+x^{-1})^2-4.$$