Question

solve it plzzz any one

Answer 1

Answer:$a {x}^{2} - 2bx = - a$

To solve this,first rationalized the denominator of x

$x = \frac{ \sqrt{ {b}^{2} + ab } + \sqrt{ {b}^{2} - ab } }{\sqrt{ {b}^{2} + ab } - \sqrt{ {b}^{2} - ab }} \times \frac{ \sqrt{ {b}^{2} + ab } + \sqrt{ {b}^{2} - ab } }{\sqrt{ {b}^{2} + ab } + \sqrt{ {b}^{2} - ab }} \\ \\ = \frac{( {\sqrt{ {b}^{2} + ab } + \sqrt{ {b}^{2} - ab }})^{2} }{ {( \sqrt{ {b}^{2} + ab } })^{2} -{( \sqrt{ {b}^{2} - ab } })^{2}} \\ \\ = \frac{ {b}^{2} + ab + {b}^{2} - ab + 2( \sqrt{{b}^{4} - {a}^{2} {b}^{2} )} }{ {b}^{2} + ab - {b}^{2} + ab } \\ \\ = \frac{2 {b}^{2} + 2b \sqrt{ {b}^{2} - {a}^{2} } }{2ab} \\ \\ x = \frac{b + \sqrt{ {b}^{2} - {a}^{2} } }{a}$
Now to solve

$a {x}^{2} - 2bx \\ \\ = x(ax - 2b) \\ \\ = \bigg(\frac{b + \sqrt{ {b}^{2} - {a}^{2} } }{a}\bigg)\bigg(a \times \frac{b + \sqrt{ {b}^{2} - {a}^{2} } }{a} - 2b\bigg) \\ \\ = \frac{b + \sqrt{ {b}^{2} - {a}^{2} } }{a}( - b + \sqrt{ {b}^{2} - {a}^{2} } ) \\ \\ = \frac{ {b}^{2} - {a}^{2} - {b}^{2} }{a} \\ \\ = \frac{ - {a}^{2} }{a} \\ \\ a {x}^{2} - 2bx = - a$
Hope it helps you.