Question

Q.15 Please...
Please answer in such way that can be understood

Answer 1

Solution:

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Given:

$a+ \sqrt{b} = c + \sqrt{d}$

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To prove:

Either a = c or b = d,

b and d are square of rationals,.

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As, we know,.

=> $a+ \sqrt{b} = c+ \sqrt{d}$

=> $c+x+ \sqrt{b} = c+ \sqrt{d}$  (for some variable x )

=> $x+ \sqrt{b} = \sqrt{d}$  .........(i)

=> $(x+ \sqrt{b} )^2 = ( \sqrt{d} )^2$

=> $x^2 + 2x \sqrt{b} +b = d$

=> $2x \sqrt{b} = d - x^2 - b$

=> $\sqrt{b} = \frac{d - x^2 - b}{2x}$

=> b = $( \frac{d - x^2 - b}{2x} )^2$

=>  ∴ b is square of rational (It can be expressed in the form a/b where b ≠ 0)

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we know that,
 
(i) => $x + \sqrt{b} = \sqrt{d}$

∴ d is square of a rational,.  (sum of two rational numbers is rational),.

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so,.

we can conclude that,

either a = c, and b = d (may be x = 0) and b & d are square of rationals,.

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                                                Hope it Helps !!