4. if x= 7+ 40, find the value of root x+ 1/root x
Answers
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Answer:
Answer:
\sqrt{x}+\frac{1}{\sqrt{x}}=\frac{4\sqrt{5}+2\sqrt{2}}{3}
x
+
x
1
=
3
4
5
+2
2
Step-by-step explanation:
i) Given x = 7+√40
=> x = 7+√2×2×10
=> x = 7+2√10
=> x = 7+2×√5×√2
=> x = 5+2+2×√5×√2
=> x = (√5)²+(√2)²+2×√5×√2
=> x = (√5+√2)²
=> √x = √5+√2 ----(1)
Now ,
ii)\frac{1}{\sqrt{x}}ii)
x
1
= \frac{1}{\sqrt{5}+\sqrt{2}}
5
+
2
1
/* Rationalising the denominator, we get
=\frac{\sqrt{5}-\sqrt{2}}{(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})}
(
5
+
2
)(
5
−
2
)
5
−
2
=\frac{\sqrt{5}-\sqrt{2}}{(\sqrt{5})^{2}-(\sqrt{2})^{2}}
(
5
)
2
−(
2
)
2
5
−
2
=\frac{\sqrt{5}-\sqrt{2}}{5-2}
5−2
5
−
2
=\frac{\sqrt{5}-\sqrt{2}}{3}
3
5
−
2
----(2)
iii) \sqrt{x}+\frac{1}{\sqrt{x}}iii)
x
+
x
1
=\sqrt{5}+\sqrt{2}+\frac{\sqrt{5}-\sqrt{2}}{3}
5
+
2
+
3
5
−
2
=\frac{3(\sqrt{5}+\sqrt{2})+\sqrt{5}-\sqrt{2}}{3}
3
3(
5
+
2
)+
5
−
2
=\frac{3\sqrt{5}+3\sqrt{2}+\sqrt{5}-\sqrt{2}}{3}
3
3
5
+3
2
+
5
−
2
=\frac{4\sqrt{5}+2\sqrt{2}}{3}
3
4
5
+2
2
Therefore,
\sqrt{x}+\frac{1}{\sqrt{x}}=\frac{4\sqrt{5}+2\sqrt{2}}{3}
x
+
x
1
=
3
4
5
+2
2