Question

Solve this quadratic equation by factorisation....

Answer 1

we can solve in this way

Answer 2

Answer:

x = -4 Or

$\implies x = \frac{9}{4}$

Explanation:

Given

$\frac{x-3}{x+3}-\frac{x+3}{x-3}=\frac{48}{7}$

Let $\frac{x-3}{x+3}=a$---(1)

Now , the equation becomes ,

$a-\frac{1}{a}=\frac{48}{7}$

$\implies \frac{(a^{2}-1)}{a}=\frac{48}{7}$

Do the cross multiplication, we get

$\implies 7(a^{2}-1)=48a$

$\implies 7a^{2}-48a-7=0$

Splitting the middle term, we get

$\implies 7a^{2}-49a+1a-7=0$

$\implies7a(a-7)+1(a-7)=0$

$\implies (a-7)(7a+1)=0$

$\implies a-7=0 \:Or \: 7a+1=0$

=> a = 7 or 7a = -1

=> a = 7 or $a=\frac{-1}{7}$

Now ,

case 1 :

If a = 7 then

$\frac{x-3}{x+3}=7$/* from (1) */

=> x-3 = 7(x+3)

=> x-3 = 7x+21

=> x-7x=21+3

=> -6x=24

$\implies x = \frac{24}{-6}$

=> x = -4

case 2:

If $a=\frac{-1}{7}$

/* from (1) */

then

$\frac{x-3}{x+3}=\frac{-1}{7}$

Do the cross multiplication,we get

=> 7(x-3) = -1(x+3)

=> 7x-21 = -x-3

=> 7x+x=-3+21

=> 8x = 18

$\implies x = \frac{18}{8}$

$\implies x = \frac{9}{4}$

Therefore,

x = -4 Or

$\implies x = \frac{9}{4}$