find a and b if √7-1/√7+1-√7+1/√7-1=a+b
Answers
Step-by-step explanation:
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
⇒ $$\dfrac{(\sqrt{7}-1)^2}{\sqrt{7}^2-1^2} }-\dfrac{(\sqrt{7}+1)^2}{\sqrt{7}^2-1^2}$$ = a + b$$\sqrt{7}$$
⇒ $$\dfrac{(\sqrt{7}-1)^2-(\sqrt{7}+1)^2}{\sqrt{7}^2-1^2} }$$ = a + b$$\sqrt{7}$$
⇒ $$\dfrac{(\sqrt{7}-1)^2-(\sqrt{7}+1)^2}{6} }$$ = a + b$$\sqrt{7}$$
Using the algebraic identity:
$$(a-b)^2$$ - $$(a+b)^2$$ = - 4ab
⇒ $$\dfrac{-4(\sqrt{7})(1)}{6} }$$ = a + b$$\sqrt{7}$$
⇒ $$\dfrac{-2\sqrt{7}}{3} }$$ = a + b$$\sqrt{7}$$
⇒ 0 + $$\sqrt{7}$$ $$(\dfrac{-2}{3} })$$ = a + b$$\sqrt{7}$$ ......... (i)
Comparing both sides, we get
a = 0 and b = $$\dfrac{-2}{3}$$
∴ a = 0 and b = $$\dfrac{-2}{3}$$
Thus, the values of a and b are "0 and $$\dfrac{-2}{3}$$ ".