root 3-1upon root3+1=a-b root3
Answers
Answer:
Answer:
Values of a = 2 and b = -1
Step-by-step explanation:
Given LHS =\frac{\sqrt{3}-1}{\sqrt{3}+1}LHS=
3
+1
3
−1
/*multiply numerator and denominator by \sqrt{3}-1
3
−1 ,we get
=\frac{(\sqrt{3}-1)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}
(
3
+1)(
3
−1)
(
3
−1)(
3
−1)
=\frac{(\sqrt{3}-1)^{2}}{(\sqrt{3})^{2}-1^{2}}
(
3
)
2
−1
2
(
3
−1)
2
/* By algebraic identity:
(a+b)(a-b) = a²-b² */
=\frac{(\sqrt{3})^{2}+1^{2}-2\times \sqrt{3}\times 1}{3-1}
3−1
(
3
)
2
+1
2
−2×
3
×1
\* By algebraic identity:
(a-b)² = a²+b²-2ab *\
=\frac{3+1-2\sqrt{3}}{2}
2
3+1−2
3
=\frac{4-2\sqrt{3}}{2}
2
4−2
3
=\frac{2(2-\sqrt{3})}{2}
2
2(2−
3
)
/* After cancellation, we get
= 2-\sqrt{3}2−
3
=2+(-1)\sqrt{3}2+(−1)
3
--(1)
= a+b\sqrt{3}a+b
3
----(2)
\* given RHS *\
Compare (1)& (2) , we get
a = 2 , b = -1
Therefore,
Values of a = 2 and b = -1
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