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+2 votes
57.9k views
in Quadratic Equations by (56.3k points)
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Find the roots of the equation \(\frac{(x-3)}{(x+3)} - \frac{(x+3)}{(x-3)} = \frac{48}{7}\), x ≠ 3, x ≠ -3.

2 Answers

+2 votes
by (30.5k points)
selected by
 
Best answer

Given equation is,

On cross-multiplying, we get

7(-12x) = 48(x2 – 9)

⇒ -84x = 48x2 – 432

⇒ 48x2 + 84x – 432 = 0

⇒ 4x2 + 7x – 36 = 0 [dividing by 12]

⇒ 4x2 + 16x – 9x – 36 = 0

⇒ 4x(x + 4) – 9(x – 4) = 0

⇒ (4x – 9)(x + 4) = 0

Now, either 4x – 9 = 0 ⇒x = \(\frac{9}{4}\)

Or, x + 4 = 0 ⇒ x = -4

Thus, the roots of the given quadratic equation are x = \(\frac{9}{4}\) and -4 respectively.

+1 vote
by (15.0k points)

We have been given

\(\frac{x-3}{x +3} - \frac{x+ 3}{x - 3} = \frac {48}7\)

7(x2 + 9 - 6x - x2 - 9 - 6x) = 48(x2 - 9)

48x2 + 84x - 432 = 0

4x2 + 7x - 36 = 0

Therefore,

4x2 + 16x - 9x - 36 = 0

4x(x + 4) - 9(x + 4) = 0

(4x - 9)(x + 4) = 0

Therefore,

4x - 9 = 0

4x = 9

x = 9/4

or,

x + 4 = 0

x = -4

Hence, x = 9/4 or x = -4.

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