then, $\frac{p^2}{q^2} + \frac{q^2}{p^2} = $
Algebraic Solution for Fraction Equation
We are given that $p$ and $q$ are positive integers and the equation:
$ \frac{p}{q} + \frac{q}{p} = 3 $
We need to find the value of:
$ \frac{p^2}{q^2} + \frac{q^2}{p^2} $
Step-by-Step Calculation
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Let $x = \frac{p}{q}$. The given equation can be rewritten in terms of $x$ as:
$ x + \frac{1}{x} = 3 $
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The expression we need to find is $\frac{p^2}{q^2} + \frac{q^2}{p^2}$, which is equivalent to $x^2 + \frac{1}{x^2}$.
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Consider the square of the term $(x + \frac{1}{x})$:
$ \left(x + \frac{1}{x}\right)^2 = x^2 + 2(x)\left(\frac{1}{x}\right) + \left(\frac{1}{x}\right)^2 $
$ \left(x + \frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2} $
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Rearrange the formula to solve for $x^2 + \frac{1}{x^2}$:
$ x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 - 2 $
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Substitute the given value $\left(x + \frac{1}{x} = 3\right)$ into the rearranged formula:
$ x^2 + \frac{1}{x^2} = (3)^2 - 2 $
$ x^2 + \frac{1}{x^2} = 9 - 2 $
$ x^2 + \frac{1}{x^2} = 7 $
-
Since $x = \frac{p}{q}$, we have $x^2 = \frac{p^2}{q^2}$ and $\frac{1}{x^2} = \frac{q^2}{p^2}$. Therefore:
$ \frac{p^2}{q^2} + \frac{q^2}{p^2} = 7 $
Conclusion
The value of $\frac{p^2}{q^2} + \frac{q^2}{p^2}$ is 7.
Important Questions from Algebra
- For positive non-zero real variables $x$ and $y$, if
$ln\left(\frac{x+y}{2}\right) = \frac{1}{2} [ln\left(x\right) + ln\left(y\right)]$
then, the value of $\frac{x}{y} + \frac{y}{x}$ is Given $f(x, y) = x^2 - 2xy + y^2$
The complete contour of the equation $f(x, y) = 1$ is described by the option(s) ___.
- It is given that $x$ and $y$ are integers in the following equation:
$$(x + y - 7)^2 + (y + 3x - 13)^2 = 0$$
The value of $(x^3 + y^3)$ is ________ (in integer). - If $pqr \neq 0$ and $p^{-x} = \frac{1}{q}$, $q^{-y} = \frac{1}{r}$, $r^{-z} = \frac{1}{p}$, what is the value of the product $xyz$?
- Two points $(4, p)$ and $(0, q)$ lie on a straight line having a slope of $3/4$. The value of $(p – q)$ is
