Find a & b if 7 + 3 root 5 / 2 + root 5 - 7 - 3 root 5/2 - root 5 = a + b root 5
Answers
Answer:
Given:
To find:
Solution:
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.