Question

solve : √x+13 + √x-1 /√x+13-√x-1 =3​

Answer 1

Given Equation

$\tt \to \: \dfrac{ \sqrt{x + 13} + \sqrt{x - 1} }{ \sqrt{x + 13} - \sqrt{x - 1} } = 3$

To find the value of x

So Take

$\tt \to \: \dfrac{ \sqrt{x + 13} + \sqrt{x - 1} }{ \sqrt{x + 13} - \sqrt{x - 1} } = 3$

Using Rationalization Method , we get

$\tt \to \: \dfrac{ \sqrt{x + 13} + \sqrt{x - 1} }{ \sqrt{x + 13} - \sqrt{x - 1} } \times\dfrac{ \sqrt{x + 13} + \sqrt{x - 1} }{ \sqrt{x + 13} + \sqrt{x - 1} } = 3$

Use this identities

$\tt \to \: (a + b)(a + b) = (a + b) {}^{2} \\ \tt \to(a + b)(a - b) = ( {a}^{2} - {b}^{2} ) \\ \tt \to(a + b) {}^{2} = {a}^{2} + {b}^{2} + 2ab \: \: \:$

we get

$\tt \to\dfrac{ (\sqrt{x + 13} + \sqrt{x - 1} ) {}^{2} }{ (\sqrt{x + 13}) {}^{2} - (\sqrt{x - 1}) {}^{2} } = 3$

$\tt \to \: \dfrac{( \sqrt{x + 13 }) {}^{2} + ( \sqrt{x - 1})^{2} + 2( \sqrt{x + 13})( \sqrt{x - 1}) }{x + 13 - x + 1} = 3$

$\tt \to \dfrac{x + 13 + x - 1 + 2 \sqrt{(x + 13)(x - 1)} }{14} = 3$

$\tt \to \: \dfrac{2x + 12 + 2 \sqrt{ {x}^{2} - x + 13x - 13 } }{14} = 3$

$\tt \to \: \dfrac{2(x + 6 + \sqrt{ {x}^{2} + 12x - 13 }) }{14} = 3$

$\tt \to \: \dfrac{x + 6 + \sqrt{ {x}^{2} + 12x - 13 } }{7} = 3$

$\tt \to \: x + 6 + \sqrt{ {x}^{2} + 12x - 13} = 21$

$\tt \to \: \sqrt{ {x}^{2} + 12x - 13 } = 21 - 6 - x$

$\tt \to \: \sqrt{ {x}^{2} + 12x - 13 } = 15- x$

Now Squaring on both side

$\tt \to \: (\sqrt{ {x}^{2} + 12x - 13 } ) {}^{2} = (15- x)^{2}$

$\tt \to \: {x}^{2} + 12x - 13 = (15- x) {}^{2}$

$\tt \to \: {x}^{2} + 12x - 13 = 225 + {x}^{2} - 30x$

$\tt \to \: 12x - 13 = 225 - 30x$

$\tt \to \: 12x + 30x= 225 + 13$

$\tt \to \: 32x = 238$

$\tt \to \: x = \dfrac{238}{32}$

$\tt \to \: x = \dfrac{119}{16}$

Answer

$\tt \to \: x = \dfrac{119}{16}$