Solve the following equation using the method | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $-2 \sqrt{2}$

Vedclass · Answer

(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 $x = -2\sqrt{2}$

Part $I$	Part $II$
$1.$ The discriminant of $x^{2}+5x+6=0$	$a. 1$
$2.$ The discriminant of $x^{2}+5x+4=0$	$b. 9$
$3.$ The discriminant of $x^{2}+4x+3=0$	$c. 4$
$4.$ The discriminant of $x^{2}+6x+5=0$	$d. 16$

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

Similar Questions

Find the roots of the following quadratic equations by the factorisation method:
$\frac{2}{5} x^{2}-x-\frac{3}{5}=0$

Solve the following quadratic equation using the method of factorization: $2x^2 + ax - a^2 = 0$.

The golden ratio,often denoted by the Greek letter $\phi$,is a special number. Which of the following is the golden number?

If the following quadratic equation has two equal and real roots,then find the value of $k$: $(k+1) x^{2}-2(k+3) x+(2 k+3)=0$

Vedclass Test Series

Exam Paper Generator

Online Exam Module