rationalize root3+root2
root3-root2
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Answered by
2
Answer:
\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=5+2\sqrt{6}
3
−
2
3
+
2
=5+2
6
Step-by-step explanation:
Given \: \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}Given
3
−
2
3
+
2
Multiply numerator and denominator by (√3+√2), we get
=\frac{(\sqrt{3}+\sqrt{2})(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}=
(
3
−
2
)(
3
+
2
)
(
3
+
2
)(
3
+
2
)
=\frac{\big(\sqrt{3}+\sqrt{2}\big)^{2}}{(\sqrt{3})^{2}-(\sqrt{2})^{2}}=
(
3
)
2
−(
2
)
2
(
3
+
2
)
2
=\frac{(\sqrt{3})^{2}+(\sqrt{2})^{2}+2\times \sqrt{3}\times \sqrt{2}}{3-2}=
3−2
(
3
)
2
+(
2
)
2
+2×
3
×
2
=\frac{3+2+2\sqrt{6}}{1}=
1
3+2+2
6
=5+2\sqrt{6}=5+2
6
•••♪
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