Question

how to solve .............................................​

Answer 1

Answer:

Step-by-step explanation:

Given :

To find the value of :

 143

   Σ $\frac{1}{\sqrt{k+1} + \sqrt{k} }$

k = 4

Solution :

$\frac{1}{\sqrt{k+1} + \sqrt{k} }$

By taking conjugate ,

⇒  $\frac{1}{\sqrt{k+1} + \sqrt{k} }$ = $\frac{1}{\sqrt{k+1}+\sqrt{k} }\times \frac{\sqrt{k+1} - \sqrt{k}}{\sqrt{k+1} - \sqrt{k}} = \frac{\sqrt{k+1} - \sqrt{k}}{(\sqrt{k+1})^2 - (\sqrt{k})^2} = \frac{\sqrt{k+1} - \sqrt{k}}{k+1 - k} = \frac{\sqrt{k+1} - \sqrt{k}}{1}$

⇒ $\sqrt{k+1} - \sqrt{k}$

⇒

  143

   Σ $\frac{1}{\sqrt{k+1} + \sqrt{k} }$

k = 4

⇒

 143

   Σ $\sqrt{k+1} - \sqrt{k}$

k = 4

⇒ $(\sqrt{4+1} - \sqrt{4} ) + (\sqrt{5+1} - \sqrt{5} ) + .... + (\sqrt{142+1} - \sqrt{142} ) + (\sqrt{143+1} - \sqrt{143} )$

⇒ $(\sqrt{5} - \sqrt{4} ) + (\sqrt{6} - \sqrt{5} ) + .... + (\sqrt{143} - \sqrt{142} ) + (\sqrt{144} - \sqrt{143} )$

⇒ $\sqrt{144} - \sqrt{4} = 12 - 2 = 10$  (not 14,.sorry)

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Given :

To find the value of :

$\frac{a + \sqrt{a^2-b^2} }{a - \sqrt{a^2-b^2}} +\frac{a - \sqrt{a^2-b^2} }{a + \sqrt{a^2-b^2}}$

Solution :

$\frac{a + \sqrt{a^2-b^2} }{a - \sqrt{a^2-b^2}} +\frac{a - \sqrt{a^2-b^2} }{a + \sqrt{a^2-b^2}}$

By cross - multiplication :

⇒ $\frac{(a + \sqrt{a^2-b^2})^2 + (a - \sqrt{a^2-b^2})^2 }{(a - \sqrt{a^2-b^2})(a + \sqrt{a^2-b^2})}$

⇒  $\frac{(a)^2 +2(a)(\sqrt{a^2-b^2})+ (\sqrt{a^2-b^2})^2 + (a)^2 - 2(a)(\sqrt{a^2-b^2})+(\sqrt{a^2-b^2})^2 }{(a)^2 - (\sqrt{a^2-b^2})^2}$

⇒ $\frac{2a^2 +2 (\sqrt{a^2-b^2})^2 }{a^2 - (\sqrt{a^2-b^2})^2}$

⇒ $\frac{2a^2 +2 (a^2-b^2) }{a^2 - (a^2-b^2)}$

⇒ $\frac{2a^2 +2a^2- 2b^2}{a^2 - a^2+b^2}$

⇒ $\frac{4a^2- 2b^2}{b^2}$

⇒ $\frac{2(2a^2- b^2)}{b^2}$

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Given :

If , $f(x)=x^2-(\sqrt{5}-1)x - (\sqrt{5}+1)$

Then find ,

$\frac{1}{\alpha^2} + \frac{1}{\beta^2 }$

Solution :

$\frac{1}{\alpha^2} + \frac{1}{\beta^2 }$

By cross-multiplication,.

⇒ $\frac{\alpha^2 +\beta^2 }{\alpha^2\beta^2 }$

⇒ $\frac{(\alpha+\beta)^2 - 2\alpha \beta }{(\alpha \beta)^2}$

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We know that,

Sum of the roots = α + β = $\frac{-b}{a} = \frac{-[-(\sqrt{5} - 1 )]}{1} = \sqrt{5} - 1$

⇒ $\alpha + \beta = \sqrt{5} - 1$

Product of the roots = αβ = $\frac{c}{a} = \frac{-(\sqrt{5} + 1 )}{1} = -\sqrt{5} - 1$

⇒ $\frac{(\alpha+\beta)^2 - 2\alpha \beta }{(\alpha \beta)^2}$

⇒ $\frac{(\sqrt{5} - 1)^2 - 2(-\sqrt{5} - 1) }{(-\sqrt{5} - 1)^2}$

⇒ $\frac{5 - 2\sqrt{5} + 1 + 2\sqrt{5} + 2 }{5 + 2\sqrt{5} + 1}$

⇒ $\frac{8}{6 + 2\sqrt{5}}$

⇒ $\frac{2(4)}{2(3 + \sqrt{5})}$

⇒ $\frac{4}{3 + \sqrt{5}}$

By taking conjugate,

⇒ $\frac{4}{3 + \sqrt{5}} \times \frac{3 - \sqrt{5}}{3 - \sqrt{5}} = \frac{4(3 - \sqrt{5})}{(3)^2 - (\sqrt{5})^2} = \frac{4(3 - \sqrt{5})}{9 - 5} = \frac{4(3 - \sqrt{5})}{4} = 3 - \sqrt{5}$