Question

if a and b are rational numbers and 4+3√5/4-3√5=a+b√5 find the value of a and b

Answer 1

Answer:

$\text{The values of a and b are 4 and }\frac{-9}{4}$

Step-by-step explanation:

Given if a and b are rational numbers and

$4+\frac{3\sqrt5}{4}-3\sqrt5=a+b\sqrt5$

$4+\frac{3\sqrt5}{4}-3\sqrt5=a+b\sqrt5\\\\\text{Taking square root of 5 common from last two terms}\\\\4+\sqrt5(\frac{3}{4}-3)=a+b\sqrt5\\\\4+\sqrt5(\frac{3-12}{4})=a+b\sqrt5\\\\4+(\frac{-9}{4})\sqrt5=a+b\sqrt5$

Comparing both sides, we get

a=4 and $b=\frac{-9}{4}$

Answer 2

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

$\tt{\rightarrow\dfrac{4+3\sqrt{5}}{4-3\sqrt{5}}}$

$\tt{\rightarrow\dfrac{4+3\sqrt{5}}{4-3\sqrt{5}}\times\dfrac{(4+3\sqrt{5})}{(4+3\sqrt{5})}}$

$\tt{\rightarrow\dfrac{(4+3\sqrt{5})^{2}}{4^2-(3\sqrt{5})^{2}}}$

$\tt{\rightarrow\dfrac{(4+3\sqrt{5})^{2}}{(16-45)}}$

$\tt{\rightarrow\dfrac{(4+3\sqrt{5})^{2}}{-29}}$

$\tt{\rightarrow\dfrac{4^2+(3\sqrt{5})^2+2\times 4\times 3\sqrt{5}}{-29}}$

$\tt{\rightarrow\dfrac{16+45+24\sqrt{5}}{-29}}$

$\tt{\rightarrow\dfrac{(61+24\sqrt{5})}{-29}}$

$\tt{\rightarrow -(\dfrac{61}{29})+(-\dfrac{24}{29})\sqrt{5}}$

$\tt{\rightarrow\dfrac{4+3\sqrt{5}}{4-3\sqrt{5}=a+b\sqrt{5}}}$

$\tt{\rightarrow -(\dfrac{61}{29})+(-\dfrac{24}{29})\sqrt{5}}$

$\tt{\rightarrow a=-\dfrac{61}{29}\;and\;b=-\dfrac{24}{29}}$