Question

If $x\ne 0$, $x+\frac{1}{2x}=p$ and $x-\frac{1}{2x}=q$; find a relation between $p$ and $q$.

• A. $p^2+q^2=2$
• B. $p^2-q^2=2$
• C. $p^3-q^3=2$
• D. $p^3-q^2=4$

Answer 1

Answer: (B)

$p^2-q^2=2$

$x+\frac{1}{2x}=p$
Squaring both the sides,
$=>(x+\frac{1}{2x}{)}^2=p^2$
$=>x^2+2\left(x\right)\left(\frac{1}{2x}\right)+(\frac{1}{2x}{)}^2=p^2$
$=>x^2+\frac{1}{4x^2}+1=p^2$ (i)
$x-\frac{1}{2x}=q$
Squaring both the sides,
$=>(x-\frac{1}{2x}{)}^2=q^2$
$=>x^2-2\left(x\right)\left(\frac{1}{2x}\right)+(\frac{1}{2x}{)}^2=q^2$
$=>x^2+\frac{1}{4x^2}-1=q^2$ (ii)
Subtractig (i) and (ii) we get,
$x^2+\frac{1}{4x^2}+1=p^2$ (i)
$x^2+\frac{1}{4x^2}-1=q^2$ (ii)

$+2=p^2-q^2$
$=>2=p^2-q^2$
$=>p^2-q^2=2$