a-b-a-b-c-a-c-a-c-b-Solve-for-real-b-and-c-in-terms-of-real-a-

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 \dots .

Answered by ajfour last updated on 18/Sep/18

\boldsymbol I f \boldsymbol a \boldsymbol i s \boldsymbol t o \boldsymbol b e \boldsymbol f o u n d \boldsymbol i n \boldsymbol t e r m s

\boldsymbol o f \boldsymbol b \boldsymbol a n d \boldsymbol 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}} \dots (\boldsymbol 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

\boldsymbol a = 1 + \frac{\boldsymbol b^{2}}{4} = 1 + \frac{\boldsymbol 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 \boldsymbol 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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