Question

Solve by quadratic equation​

Answer 1

Question :

• $\tt \dfrac{x+2}{x-2} + \dfrac{x-2}{x+2} = \dfrac{5}{6}$

Answer :

• $\large \underline{\boxed{\bf{x = \dfrac{\pm 2\sqrt{119}}{7}}}}$

Explanation :

$\tt : \implies \dfrac{x+2}{x-2} + \dfrac{x-2}{x+2} = \dfrac{5}{6}$

$\tt : \implies \dfrac{(x+2)(x+2) + (x-2)(x-2)}{(x-2)(x+2)} = \dfrac{5}{6}$

$\tt : \implies \dfrac{(x+2)^{2} + (x-2)^{2}}{(x-2)(x+2)} = \dfrac{5}{6}$

Now, we know that :

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

$\tt : \implies \dfrac{x^{2}+2^{2}+ 2 \times x \times 2 + x^{2}+2^{2} - 2 \times x \times 2 }{x^{2}-2^{2}} = \dfrac{5}{6}$

$\tt : \implies \dfrac{x^{2}+ 4 + \cancel{4x} + x^{2}+ 4 - \cancel{4x}}{x^{2}-4} = \dfrac{5}{6}$

$\tt : \implies \dfrac{x^{2} + x^{2} + 4 + 4}{x^{2}-4} = \dfrac{5}{6}$

$\tt : \implies \dfrac{2x^{2} + 8}{x^{2}-4} = \dfrac{5}{6}$

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 :

$\large \underline{\boxed{\bf{x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}}}}$

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

$\tt : \implies x = \dfrac{0 \pm \sqrt{0 - 1904}}{14}$

$\tt : \implies x = \dfrac{\pm \sqrt{- 1904}}{14}$

$\tt : \implies x = \dfrac{\pm \sqrt{2\times 2\times 2\times 2\times 7\times 17}}{14}$

$\tt : \implies x = \dfrac{\pm \cancel{2} \times 2\sqrt{7\times 17}}{\cancel{14}}$

$\tt : \implies x = \dfrac{\pm2\sqrt{119}}{7}$

$\large \underline{\boxed{\bf{x = \dfrac{\pm 2\sqrt{119}}{7}}}}$

Hence value of $\bf x =\dfrac{\pm 2\sqrt{119}}{7}$

Answer 2

Answer:

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