Find a and b from the above question. Please give step by step explanation also.
Answers
Step-by-step explanation:
Answer:
so \: the \: values \: are \: \boxed{a = 0} \: and \: \boxed{b = 1}sothevaluesare
a=0
and
b=1
Step-by-step explanation:
\frac{7 + \sqrt{5} }{7 - \sqrt{5} } - \frac{7 - \sqrt{5} }{7 + \sqrt{5} } = a + b\frac{7}{11}\sqrt{5}
7−
5
7+
5
−
7+
5
7−
5
=a+b
11
7
5
\frac{(7 + \sqrt{5} )(7 + \sqrt{5} )}{(7 - \sqrt{5})(7 + \sqrt{5} ) } - \frac{(7 - \sqrt{5} )(7 - \sqrt{5} )}{(7 + \sqrt{5} )(7 - \sqrt{5} )} = a + b\frac{7}{11} \sqrt{5}
(7−
5
)(7+
5
)
(7+
5
)(7+
5
)
−
(7+
5
)(7−
5
)
(7−
5
)(7−
5
)
=a+b
11
7
5
\frac{(7 + \sqrt{5} )^{2} }{ {(7)}^{2} - {( \sqrt{5} )}^{2} } - \frac{(7 - \sqrt{5} )^{2} }{ {(7)}^{2} - {( \sqrt{5} )}^{2} } = a + b\frac{7}{11} \sqrt{5}
(7)
2
−(
5
)
2
(7+
5
)
2
−
(7)
2
−(
5
)
2
(7−
5
)
2
=a+b
11
7
5
\frac{(7)^{2} + 2 \times 7 \times \sqrt{5} + {( \sqrt{5} )}^{2} }{49 - 5} - \frac{(7)^{2} - 2 \times 7 \times \sqrt{5} }{49 - 5} = a + b \frac{7}{11} \sqrt{5}
49−5
(7)
2
+2×7×
5
+(
5
)
2
−
49−5
(7)
2
−2×7×
5
=a+b
11
7
5
\frac{49 + 5 + 14 \sqrt{5} }{44} - \frac{49 + 5 - 14 \sqrt{5}}{44} = a + b \frac{7}{11} \sqrt{5}
44
49+5+14
5
−
44
49+5−14
5
=a+b
11
7
5
\frac{54 + 14 \sqrt{5} - (54 - 14 \sqrt{5} ) }{44} = a + b \frac{7}{11} \sqrt{5}
44
54+14
5
−(54−14
5
)
=a+b
11
7
5
\frac{54 + 14 \sqrt{5} - 54 + 14 \sqrt{5} }{44} = a + b \frac{7}{11} \sqrt{5}
44
54+14
5
−54+14
5
=a+b
11
7
5
\frac{28 \sqrt{5} }{44} = a + b \frac{7}{11} \sqrt{5}
44
28
5
=a+b
11
7
5
\frac{7}{11} \sqrt{5} = a + b \frac{7}{11} \sqrt{5}
11
7
5
=a+b
11
7
5
by \: comparision \: of \: both \: sides \: we \: getbycomparisionofbothsidesweget
a = 0 \: and \: b = 1a=0andb=1