If a = 7-4/3 , then the value of a +1/a is__
Answers
Answer:
The given value of a is 7-4 \sqrt{3}7−4
3
And the value of \frac{1}{a} \text { is } 7+4 \sqrt{3}
a
1
is 7+4
3
Then,
\begin{gathered}\begin{array}{c}{a+\frac{1}{a}=7-4 \sqrt{3}+7+4 \sqrt{3}} \\ {a+\frac{1}{a}=14}\end{array}\end{gathered}
a+
a
1
=7−4
3
+7+4
3
a+
a
1
=14
Now to find the value of \sqrt{a}+\frac{1}{\sqrt{a}}
a
+
a
1
,
By using the formula of (a+b)^{2}=a^{2}+b^{2}+2 a b(a+b)
2
=a
2
+b
2
+2ab
Then, applying the (a+b)^{2}(a+b)
2
formula to \sqrt{a}+\frac{1}{\sqrt{a}}
a
+
a
1
We can get,
\begin{gathered}\begin{array}{c}{\left(\sqrt{a}+\frac{1}{\sqrt{a}}\right)^{2}=(\sqrt{a})^{2}+\left(\frac{1}{\sqrt{a}}\right)^{2}+\left(2 \times \sqrt{a} \times \frac{1}{\sqrt{a}}\right)} \\ {\left(\sqrt{a}+\frac{1}{\sqrt{a}}\right)^{2}=a+\frac{1}{a}+2}\end{array}\end{gathered}
(
a
+
a
1
)
2
=(
a
)
2
+(
a
1
)
2
+(2×
a
×
a
1
)
(
a
+
a
1
)
2
=a+
a
1
+2
Already we know the value of a+\frac{1}{a}a+
a
1
is 14
\begin{gathered}\begin{array}{c}{\left(\sqrt{a}+\frac{1}{\sqrt{a}}\right)^{2}=14+2} \\ {\left(\sqrt{a}+\frac{1}{\sqrt{a}}\right)^{2}=16}\end{array}\end{gathered}
(
a
+
a
1
)
2
=14+2
(
a
+
a
1
)
2
=16
By taking square root for the value of \left(\sqrt{a}+\frac{1}{\sqrt{a}}\right)^{2}(
a
+
a
1
)
2
, we can get the value of \sqrt{a}+\frac{1}{\sqrt{a}}
a
+
a
1
\begin{gathered}\begin{array}{l}{\sqrt{a}+\frac{1}{\sqrt{a}}=\sqrt{16}} \\ {\sqrt{a}+\frac{1}{\sqrt{a}}=4}\end{array}\end{gathered}
a
+
a
1
=
16
a
+
a
1
=4
Then the value of \bold{\sqrt{a}+\frac{1}{\sqrt{a}}}
a
+
a
1
is 4, if the value of a is 7-4 \sqrt{3}7−4
3
.
Step-by-step explanation:
a = 7-4/3
a = 3/3 = 1
So, a=1
a+1/a
= 1+1/1
=2/1
=2