Question 89

If $$\frac{p}{q}=\frac{x+3}{x-3}$$, then what is the value of $$\frac{p^2+q^2}{p^2-q^2}$$ ?

Solution

Given : $$\frac{p}{q}=\frac{x+3}{x-3}$$

Let $$p=(x+3)$$ and $$q=(x-3)$$

To find : $$\frac{p^2+q^2}{p^2-q^2}$$

= $$\frac{(x+3)^2+(x-3)^2}{(x+3)^2-(x-3)^2}$$

= $$\frac{(x^2+6x+9)+(x^2-6x+9)}{(x^2+6x+9)-(x^2-6x+9)}$$

= $$\frac{2x^2+18}{12x}$$

= $$\frac{x^2+9}{6x}$$

=> Ans - (D)

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