d) If (a +1/2 ) = 6, find a? + 1 a2
Answers
Answer:
Here is your answer...✌
GIVEN THAT,
\begin{gathered}a + \frac{1}{a} = 6 \\ \\ Squaring \: both \: sides \\ we \:get \\ \\ {(a + \frac{1}{a} ) }^{2} = {6}^{2} \\ \\ {a}^{2} + { \frac{1}{a {}^{2} } } + 2 \times a \times \frac{1}{a} = 36 \\ \\ {a}^{2} + \frac{1}{ {a}^{2} } + 2 = 36 \\ \\ {a}^{2} + \frac{1}{ {a}^{2} } = 34 \\ \\ Subtracting \: \: 2\: \: both \: \: side s \\ we \: get \\ \\ {a}^{2} + \frac{1}{ {a}^{2} } - 2 = 34 - 2 \\ \\ {a}^{2} + \frac{1}{ {a}^{2} } - 2 \times a \times \frac{1}{a} = 32 \\ \\ {(a - \frac{1}{a} ) }^{2} = 32 \\ \\ {(a - \frac{1}{a} ) }= \sqrt{32} \:\: or \: \: -\sqrt{32} \\ \\ {(a - \frac{1}{a} ) }= 4\sqrt{2} \: \: or \: \: - 4 \sqrt{2} \\ \\Now, \\\\ {a}^{2} - \frac{1}{ {a}^{2} } = (a + \frac{1}{a} )(a - \frac{1}{a} ) \\ \\ when \: (a - \frac{1}{a} ) = 4 \sqrt{2 } \\ \\ (a {}^{2} - \frac{1}{a {}^{2} } ) = 6 \times 4 \sqrt{2} \\ \\ = 24 \sqrt{2} \\ \\ when \: (a - \frac{1}{a} ) = - 4 \sqrt{2} \\ \\ (a {}^{2} - \frac{1}{a {}^{2} } ) = 6 \times ( - 4 \sqrt{2} ) \\ \\ = - 24 \sqrt{2} \end{gathered}
a+
a
1
=6
Squaringbothsides
weget
(a+
a
1
)
2
=6
2
a
2
+
a
2
1
+2×a×
a
1
=36
a
2
+
a
2
1
+2=36
a
2
+
a
2
1
=34
Subtracting2bothsides
weget
a
2
+
a
2
1
−2=34−2
a
2
+
a
2
1
−2×a×
a
1
=32
(a−
a
1
)
2
=32
(a−
a
1
)=
32
or−
32
(a−
a
1
)=4
2
or−4
2
Now,
a
2
−
a
2
1
=(a+
a
1
)(a−
a
1
)
when(a−
a
1
)=4
2
(a
2
−
a
2
1
)=6×4
2
=24
2
when(a−
a
1
)=−4
2
(a
2
−
a
2
1
)=6×(−4
2
)
=−24
2