Step-by-step explanation: The value of a is 0 and b is 2. Given: \sf{\dfrac{7+3\sqrt5}{2+\sqrt5}-\dfrac{7-3\sqrt5}{2-\sqrt5}=a+b\sqrt5} 2+ 5 7+3 5 − 2− 5 7−3 5 =a+b 5 To find: \sf{The \ value \ of \ a \ and \ b.}The value of a and b. Solution: \sf{\leadsto{a+b\sqrt5=\dfrac{7+3\sqrt5}{2+\sqrt5}-\dfrac{7-3\sqrt5}{2-\sqrt5}}}⇝a+b 5 = 2+ 5 7+3 5 − 2− 5 7−3 5 \sf{\therefore{a+b\sqrt5=\dfrac{(7+3\sqrt5)(2-\sqrt5)}{(2+\sqrt5)(2-\sqrt5)}-\dfrac{(7-3\sqrt5)(2+\sqrt5)}{(2-\sqrt5)(2+\sqrt5)}}}∴a+b 5 = (2+ 5 )(2− 5 ) (7+3 5 )(2− 5 ) − (2− 5 )(2+ 5 ) (7−3 5 )(2+ 5 ) \sf{\therefore{a+b\sqrt5=\dfrac{14-7\sqrt5+6\sqrt5-15}{2^{2}-\sqrt5^{2}}-\dfrac{14+7\sqrt5-6\sqrt5-15}{2^{2}-\sqrt5^{2}}}}∴a+b 5 = 2 2 − 5 2 14−7 5 +6 5 −15 − 2 2 − 5 2 14+7 5 −6 5 −15 \sf{\therefore{a+b\sqrt5=\dfrac{-1-\sqrt5}{-1}-\dfrac{-1+\sqrt5}{-1}}}∴a+b 5 = −1 −1− 5 − −1 −1+ 5 \sf{\therefore{a+b\sqrt5=1+\sqrt5-(1-\sqrt5)}}∴a+b 5 =1+ 5 −(1− 5 ) \sf{\therefore{a+b\sqrt5=1+\sqrt5-1+\sqrt5}}∴a+b 5 =1+ 5 −1+ 5 \sf{\therefore{a+b\sqrt5=0+2\sqrt5}}∴a+b 5 =0+2 5 \sf{On \ comparing \ we \ get,}On comparing we get, \sf{a=0 \ and \ b\sqrt5=2\sqrt5}a=0 and b 5 =2 5 \sf{\therefore{a=0 \ and \ b=2}}∴a=0 and b=2 \sf\purple{\tt{\therefore{The \ value \ of \ a \ is \ 0 \ and \ b \ is \ 2.}}}∴The value of a is 0 and b is 2. Reply
Step-by-step explanation:
The value of a is 0 and b is 2.
Given:
\sf{\dfrac{7+3\sqrt5}{2+\sqrt5}-\dfrac{7-3\sqrt5}{2-\sqrt5}=a+b\sqrt5}
2+
5
7+3
5
−
2−
5
7−3
5
=a+b
5
To find:
\sf{The \ value \ of \ a \ and \ b.}The value of a and b.
Solution:
\sf{\leadsto{a+b\sqrt5=\dfrac{7+3\sqrt5}{2+\sqrt5}-\dfrac{7-3\sqrt5}{2-\sqrt5}}}⇝a+b
5
=
2+
5
7+3
5
−
2−
5
7−3
5
\sf{\therefore{a+b\sqrt5=\dfrac{(7+3\sqrt5)(2-\sqrt5)}{(2+\sqrt5)(2-\sqrt5)}-\dfrac{(7-3\sqrt5)(2+\sqrt5)}{(2-\sqrt5)(2+\sqrt5)}}}∴a+b
5
=
(2+
5
)(2−
5
)
(7+3
5
)(2−
5
)
−
(2−
5
)(2+
5
)
(7−3
5
)(2+
5
)
\sf{\therefore{a+b\sqrt5=\dfrac{14-7\sqrt5+6\sqrt5-15}{2^{2}-\sqrt5^{2}}-\dfrac{14+7\sqrt5-6\sqrt5-15}{2^{2}-\sqrt5^{2}}}}∴a+b
5
=
2
2
−
5
2
14−7
5
+6
5
−15
−
2
2
−
5
2
14+7
5
−6
5
−15
\sf{\therefore{a+b\sqrt5=\dfrac{-1-\sqrt5}{-1}-\dfrac{-1+\sqrt5}{-1}}}∴a+b
5
=
−1
−1−
5
−
−1
−1+
5
\sf{\therefore{a+b\sqrt5=1+\sqrt5-(1-\sqrt5)}}∴a+b
5
=1+
5
−(1−
5
)
\sf{\therefore{a+b\sqrt5=1+\sqrt5-1+\sqrt5}}∴a+b
5
=1+
5
−1+
5
\sf{\therefore{a+b\sqrt5=0+2\sqrt5}}∴a+b
5
=0+2
5
\sf{On \ comparing \ we \ get,}On comparing we get,
\sf{a=0 \ and \ b\sqrt5=2\sqrt5}a=0 and b
5
=2
5
\sf{\therefore{a=0 \ and \ b=2}}∴a=0 and b=2
\sf\purple{\tt{\therefore{The \ value \ of \ a \ is \ 0 \ and \ b \ is \ 2.}}}∴The value of a is 0 and b is 2.