Question

Solve by quadratic equation​

Answer 1

I think this answer is right if no tell me please :)

Answer 2

Question :

• 
  `\tt (x+2)/(x-2) + (x-2)/(x+2) = (5)/(6)`

Answer :

• 
  `\large \underline{\boxed{\bf{x = (\pm 2√(119))/(7)}}}`

Explanation :

`\tt : \implies (x+2)/(x-2) + (x-2)/(x+2) = (5)/(6)`

`\tt : \implies ((x+2)(x+2) + (x-2)(x-2))/((x-2)(x+2)) = (5)/(6)`

`\tt : \implies ((x+2)^(2) + (x-2)^(2))/((x-2)(x+2)) = (5)/(6)`

Now, we know that :

• 
  `\large \underline{\boxed{\bf{(a+b)^(2) = a^(2) + b^(2)+ 2ab}}}`
• 
  `\large \underline{\boxed{\bf{(a-b)^(2) = a^(2) + b^(2) - 2ab}}}`
• 
  `\large \underline{\boxed{\bf{(a+b)(a-b) = a^(2) - b^(2)}}}`

`\tt : \implies (x^(2)+2^(2)+ 2 * x * 2 + x^(2)+2^(2) - 2 * x * 2 )/(x^(2)-2^(2)) = (5)/(6)`

`\tt : \implies \frac{x^(2)+ 4 + \cancel{4x} + x^(2)+ 4 - \cancel{4x}}{x^(2)-4} = (5)/(6)`

`\tt : \implies (x^(2) + x^(2) + 4 + 4)/(x^(2)-4) = (5)/(6)`

`\tt : \implies (2x^(2) + 8)/(x^(2)-4) = (5)/(6)`

By cross multiply :

`\tt : \implies (2x^(2) + 8)6= 5(x^(2)-4)`

`\tt : \implies 12x^(2) + 48 = 5x^(2)-20`

`\tt : \implies 12x^(2) + 48 - 5x^(2) + 20 = 0`

`\tt : \implies 7x^(2) + 68 = 0`

`\tt : \implies 7x^(2) + 0x + 68 = 0`

Now, by comparing with ax² + bx + c = 0, we have :

• a = 7
• b = 0
• c = 68

By using quadratic formula :

`\large \underline{\boxed{\bf{x = \frac{-b \pm \sqrt{b^(2) - 4ac}}{2a}}}}`

`\tt : \implies x = \frac{-(0) \pm \sqrt{(0)^(2) - 4(7)(68)}}{2(7)}`

`\tt : \implies x = (0 \pm √(0 - 1904))/(14)`

`\tt : \implies x = (\pm √(- 1904))/(14)`

`\tt : \implies x = (\pm √(2* 2* 2* 2* 7* 17))/(14)`

`\tt : \implies x = \frac{\pm \cancel{2} * 2√(7* 17)}{\cancel{14}}`

`\tt : \implies x = (\pm2√(119))/(7)`

`\large \underline{\boxed{\bf{x = (\pm 2√(119))/(7)}}}`

Hence value of
`\bf x =(\pm 2√(119))/(7)`