Question

√7+1/√7+1-√7+1/√7-1=a+b√7 find the value of a and b

Answer 1

We have:

$\dfrac{\sqrt{7}-1}{\sqrt{7}+1} -\dfrac{\sqrt{7}+1}{\sqrt{7}-1}$ = a + b$\sqrt{7}$

We have to find, the values of a and b are:

Solution:

∴ $\dfrac{\sqrt{7}-1}{\sqrt{7}+1} -\dfrac{\sqrt{7}+1}{\sqrt{7}-1}$ = a + b$\sqrt{7}$

Rationalising numerator and denominator, we get

$\dfrac{\sqrt{7}-1}{\sqrt{7}+1}\times \dfrac{\sqrt{7}-1}{\sqrt{7}-1} -\dfrac{\sqrt{7}+1}{\sqrt{7}-1}\times \dfrac{\sqrt{7}+1}{\sqrt{7}+1}$ = a + b$\sqrt{7}$

Using the algebraic identity:

(a + b)(a - b) = $a^{2} -b^{2}$

⇒ $\dfrac{(\sqrt{7}-1)^2}{\sqrt{7}^2-1^2} }-\dfrac{(\sqrt{7}+1)^2}{\sqrt{7}^2-1^2}$ = 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 $\dfrac{-2}{3}$".