Question

Find the value of a and b
4+3√5÷4-3√5 =a+b√5​

Answer 1

[tex]\mathfrak{\huge{Answer:-}} \\ \bold{\frac{4 + 3 \sqrt{5} }{4 - 3 \sqrt{5} } = a + b \sqrt{5}} \\ \frac{4 + 3 \sqrt{5} }{4 - 3 \sqrt{5} } \times \frac{4 + 3 \sqrt{5} }{4 + 3 \sqrt{5} } = a + b \sqrt{5} \\ \frac{(4 + 3 \sqrt{5})(4 + 3 \sqrt{5} ) }{(4 - 3 \sqrt{5} )(4 + 3 \sqrt{5}) } = a + b \sqrt{5} \\ \frac{ {(4 + 3 \sqrt{5}) }^{2} }{ {(4)}^{2} - {(3 \sqrt{5}) }^{2} } = a + b \sqrt{5} \\ \frac{16 + 45 + 24 \sqrt{5} }{16 - 45} = a + b \sqrt{5} \\ \frac{61 + 24 \sqrt{5} }{( - 29)}= a + b \sqrt{5} \\ \frac{( - 61)}{29} + \frac{( - 24 \sqrt{5} )}{29} = a + b \sqrt{5} \\ a = \frac{( - 61)}{29}, \: b = \frac{( - 24)}{29} \\ I \: am \: not \: sure \: whether \: the \: answer \: is \: correct .[/tex]