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(D) Given equation: $\frac{x^{2}-1}{x^{2}+1}=\frac{7}{9}$Cross-multiplying,we get:$9(x^{2}-1) = 7(x^{2}+1)$Expanding the terms:$9x^{2} - 9 = 7x^{2} +

Vedclass · Accepted Answer

$2 \sqrt{2}$ and $-2 \sqrt{2}$

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(D) Given equation: $\frac{x^{2}-1}{x^{2}+1}=\frac{7}{9}$Cross-multiplying,we get:$9(x^{2}-1) = 7(x^{2}+1)$Expanding the terms:$9x^{2} - 9 = 7x^{2} + 7$Rearranging the terms to one side:$9x^{2} - 7x^{2} - 9 - 7 = 0$Simplifying the equation:$2x^{2} - 16 = 0$Dividing by $2$:$x^{2} - 8 = 0$Using the difference of squares identity $a^{2}-b^{2}=(a-b)(a+b)$,where $8 = (\sqrt{8})^{2} = (2\sqrt{2})^{2}$:$(x - 2\sqrt{2})(x + 2\sqrt{2}) = 0$Setting each factor to zero:$x - 2\sqrt{2} = 0$ or $x + 2\sqrt{2} = 0$Therefore,the roots are:$x = 2\sqrt{2}$ or $x = -2\sqrt{2}$

Solve the following equation using the method of factorization: $\frac{x^{2}-1}{x^{2}+1}=\frac{7}{9}$

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