If √7-1/√7+1 - √7+1/√7-1 = a+b√7, find the value of a and b
Answers
Answer:
We have:
\dfrac{\sqrt{7}-1}{\sqrt{7}+1} -\dfrac{\sqrt{7}+1}{\sqrt{7}-1}
7
+1
7
−1
−
7
−1
7
+1
= a + b\sqrt{7}
7
We have to find, the values of a and b are:
Solution:
∴ \dfrac{\sqrt{7}-1}{\sqrt{7}+1} -\dfrac{\sqrt{7}+1}{\sqrt{7}-1}
7
+1
7
−1
−
7
−1
7
+1
= a + b\sqrt{7}
7
Rationalising numerator and denominator, we get
\dfrac{\sqrt{7}-1}{\sqrt{7}+1}\times \dfrac{\sqrt{7}-1}{\sqrt{7}-1} -\dfrac{\sqrt{7}+1}{\sqrt{7}-1}\times \dfrac{\sqrt{7}+1}{\sqrt{7}+1}
7
+1
7
−1
×
7
−1
7
−1
−
7
−1
7
+1
×
7
+1
7
+1
= a + b\sqrt{7}
7
Using the algebraic identity:
(a + b)(a - b) = a^{2} -b^{2}a
2
−b
2
⇒ = a + b\sqrt{7}
7
⇒ = a + b\sqrt{7}
7
⇒ = a + b\sqrt{7}
7
Using the algebraic identity:
(a-b)^2(a−b)
2
- (a+b)^2(a+b)
2
= - 4ab
⇒ = a + b\sqrt{7}
7
⇒ = a + b\sqrt{7}
7
⇒ 0 + \sqrt{7}
7
= a + b\sqrt{7}
7
......... (i)
Comparing both sides, we get
a = 0 and b = \dfrac{-2}{3}
3
−2
∴ a = 0 and b = \dfrac{-2}{3}
3
−2
Thus, the values of a and b are "0 and \dfrac{-2}{3}
3
−2
".