(√(a−b)) + (√(a+b)) = c (√(a−c)) + (√(a+c)) = b Solve for real b, and c ; in terms of real a.


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Question Number 43902 by ajfour last updated on 17/Sep/18

$\sqrt{a - b} + \sqrt{a + b} = c$ $\sqrt{a - c} + \sqrt{a + c} = b$ $S o l v e f o r r e a l b, a n d c; i n t e r m s$ $o f r e a l a .$
Commented by MrW3 last updated on 17/Sep/18

$a ⩾ 2$ $b = c = 2 \sqrt{a - 1}$
Commented by ajfour last updated on 17/Sep/18

$T h i s i s c e r t a i n l y r i g h t s i r; n o$ $o t h e r r e a l a n s w e r s e v e n, S i r ?$
Commented by LYCON TRIX last updated on 17/Sep/18

$I^{'} m impressed to see this question$ $I^{'} ll make this question featured in NS 7 UC$
Commented by LYCON TRIX last updated on 17/Sep/18

$what if I said that$ $Solve for a in terms of b and c$ $(a, b, c) \in R$
	Answered by tanmay.chaudhury50@gmail.com last updated on 17/Sep/18

	$b = a c o s 2 α c = a c o s 2 β$ $\sqrt{2 a} s i n α + \sqrt{2 a} c o s α = a c o s 2 β$ $\frac{s i n α + c o s α}{c o s 2 β} . = \frac{\sqrt{a}}{\sqrt{2}}$ $\frac{s i n β + c o s β}{c o s 2 α} = \frac{\sqrt{a}}{\sqrt{2}}$ $\frac{s i n α + c o s α}{c o s 2 β} = \frac{s i n β + c o s β}{c o s 2 α}$ $\sqrt{1 + s i n 2 α} c o s 2 α = \sqrt{1 + s i n 2 β} c o s 2 β$ $\sqrt{1 + \frac{2 t_{1}}{1 + t_{1}^{2}}} \times \frac{1 - t_{1}^{2}}{1 + t_{1}^{2}} = \sqrt{1 + \frac{2 t_{2}}{1 + t_{2}^{2}}} \times \frac{1 - t_{2}^{2}}{1 + t_{2}^{2}}$ $\frac{1 + t_{1} \times 1 - t_{1}^{2}}{{(1 + t_{1}^{2})}^{\frac{3}{2}}} = \frac{1 + t_{2}^{2} \times 1 - t_{2}^{2}}{{(1 + t_{2})}^{\frac{3}{2}}}$ $\frac{{(1 + t_{1})}^{2} \times (1 - t_{1})}{{(1 + t_{1}^{2})}^{\frac{3}{2}}} = \frac{{(1 + t_{2}^{})}^{2} \times (1 - t_{2})}{{(1 + t_{2})}^{\frac{3}{2}}}$ ${(\frac{1 + t_{1}}{1 + t_{2}})}^{2} \times \frac{1 - t_{1}}{1 - t_{2}} = {(\frac{1 + t_{1}^{2}}{1 + t_{2}^{2}})}^{\frac{3}{2}}$ $t_{1} = t a n α t_{2} = t a n β$ $c o n t d . . . .$
	Answered by ajfour last updated on 18/Sep/18

	$I f a i s t o b e f o u n d i n t e r m s$ $o f b a n d c .$ $s q u a r i n g f i r s t e q .$ $2 a + 2 \sqrt{a^{2} - b^{2}} = c^{2}$ $s q u a r i n g a g a i n$ $4 a^{2} - 4 b^{2} = c^{4} + 4 a^{2} - 4 {a c}^{2}$ $\Rightarrow a = \frac{c^{2}}{4} + \frac{b^{2}}{c^{2}}$ $s i m i l a r l y f r o m s e c o n d e q .$ $a = \frac{b^{2}}{4} + \frac{c^{2}}{b^{2}}$ $\Rightarrow a = \frac{c^{2}}{4} + \frac{b^{2}}{c^{2}} = \frac{b^{2}}{4} + \frac{c^{2}}{b^{2}} . . . (i)$ $\Rightarrow \frac{c^{2} - b^{2}}{4} = \frac{(c^{2} - b^{2}) (c^{2} + b^{2})}{b^{2} c^{2}}$ $T w o c a s e s a r i s e :$ $I f b^{2} = c^{2}, t h e n$ $a = 1 + \frac{b^{2}}{4} = 1 + \frac{c^{2}}{4}$ $I f b^{2} \neq c^{2}$ $\Rightarrow b^{2} + c^{2} = \frac{b^{2} c^{2}}{4}, t h e n u s i n g (i)$ $2 a = \frac{b^{2} + c^{2}}{4} + \frac{b^{4} + c^{4}}{b^{2} c^{2}}$ $\Rightarrow 2 a = \frac{b^{2} + c^{2}}{4} + \frac{{(b^{2} + c^{2})}^{2} - 2 b^{2} c^{2}}{b^{2} c^{2}}$ $2 a = \frac{b^{2} c^{2}}{16} + \frac{b^{4} c^{4}}{16 b^{2} c^{2}} - 2$ $\Rightarrow a = \frac{b^{2} c^{2}}{16} - 1 = \frac{b^{2} + c^{2}}{4} - 1 .$
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