Solve-39-10-a-a-2-9-a-2-2-14-a-R-Mr-Rasheed-yesterday-i-solved-it-on-a-draft-paper-When-i-came-today-to-write-it-down-on-the-notebook-i-got-into-a-maze-do-you

Tinku Tara June 4, 2023 Algebra 0 Comments

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Question Number 182681 by Acem last updated on 12/Dec/22

Solve (√(39− 10 a −a^2 )) + (√(9−a^2 )) = 2 (√(14)) ; a∈ R^(+★) @Mr. Rasheed, yesterday i solved it on a draft paper When i came today to write it down on the notebook i got into a maze, do you have an easy way to deal with? “if you have time” i lost that paper, and this formula makes me laugh at myself, it′s for 13 y/o “confused face” Sure Every friend can share!

S o l v e \sqrt{39 - 10 a - a^{2}} + \sqrt{9 - a^{2}} = 2 \sqrt{14}; a \in R^{+ ★}

@ M r . R a s h e e d, y e s t e r d a y i s o l v e d i t o n a d r a f t p a p e r

W h e n i c a m e t o d a y t o w r i t e i t d o w n o n

t h e n o t e b o o k i g o t i n t o a m a z e, d o y o u h a v e

a n e a s y w a y t o d e a l w i t h ? “ i f y o u h a v e {t i m e}^{″}

i l o s t t h a t p a p e r, a n d t h i s f o r m u l a m a k e s m e

l a u g h a t m y s e l f, {i t}^{'} s f o r 13 y / o “ c o n f u s e d {f a c e}^{″}

S u r e E v e r y f r i e n d c a n s h a r e!

Commented by mr W last updated on 12/Dec/22

2(√(14)) or 2(√(24)) ?

2 \sqrt{14} o r 2 \sqrt{24} ?

Commented by Acem last updated on 12/Dec/22

No, it′s about 1.383 8 Sir , am not ok right now i will solve it again later though it′s so boring

N o, {i t}^{'} s a b o u t 1.383 8 S i r, a m n o t o k r i g h t n o w

i w i l l s o l v e i t a g a i n l a t e r t h o u g h {i t}^{'} s s o b o r i n g

Commented by mr W last updated on 13/Dec/22

i didn′t mean the answer is 2(√(14)) or 2(√(24)). i asked if it is 2(√(14)) or 2(√(24)) in the question. i have a different solution than you all, see below.

i {d i d n}^{'} t m e a n t h e a n s w e r i s 2 \sqrt{14} o r

2 \sqrt{24} . i a s k e d i f i t i s 2 \sqrt{14} o r 2 \sqrt{24} i n t h e

q u e s t i o n .

i h a v e a d i f f e r e n t s o l u t i o n t h a n y o u

a l l, s e e b e l o w .

Commented by manxsol last updated on 13/Dec/22

Sir W could share his solution.I would appreciate it if you critque my solution.

Sir W could share his solution . I

would appreciate it if you

critque my solution .

Commented by manxsol last updated on 14/Dec/22

combination metod friz+metod rasheed solution (√x)+(√y)=(√u) x(a),y(a), u=k 4ux=[u+(x−y)]^2 I 4uy=[u−(x−y)]^2 II sea x−y=qa+b lineal solve I ⇒ma^2 +na+p solve II⇒ ma^2 +na+p solve ma^2 +na+p=0 a_1 a_2 DEVELOPING (√x)+(√y)=(√u) ((√x)−(√y))((√x)+(√y))=(√u)((√x)−(√y)) x−y=(√u) ((√x)−(√y)) .................. (√x)−(√y)=(((x−y))/( (√u))) (√x)+(√y)=(√u) .................. add 2(√x)=(((u+(x−y)))/( (√u))) 4xu=(u+(x−y)^2 ) 4yu=(u−(x−y)^2 ) CONCLUSION (√x)+(√y)=(√u) is equivalent a solve {4xu=(u+(x−y))^2 } ∪ {4yu=(u−(x−y))^(2 ) }

c o m b i n a t i o n

m e t o d f r i z + m e t o d r a s h e e d

s o l u t i o n

\sqrt{x} + \sqrt{y} = \sqrt{u} x (a), y (a), u = k

4 u x = {[u + (x - y)]}^{2} I

4 u y = {[u - (x - y)]}^{2} I I

s e a x - y = q a + b l i n e a l

s o l v e I \Rightarrow {m a}^{2} + n a + p

s o l v e I I \Rightarrow {m a}^{2} + n a + p

s o l v e {m a}^{2} + n a + p = 0

a_{1}

a_{2}

D E V E L O P I N G

\sqrt{x} + \sqrt{y} = \sqrt{u}

(\sqrt{x} - \sqrt{y}) (\sqrt{x} + \sqrt{y}) = \sqrt{u} (\sqrt{x} - \sqrt{y})

x - y = \sqrt{u} (\sqrt{x} - \sqrt{y})

\dots \dots \dots \dots \dots \dots

\sqrt{x} - \sqrt{y} = \frac{(x - y)}{\sqrt{u}}

\sqrt{x} + \sqrt{y} = \sqrt{u}

\dots \dots \dots \dots \dots \dots

a d d

2 \sqrt{x} = \frac{(u + (x - y))}{\sqrt{u}}

4 x u = (u + {(x - y)}^{2})

4 y u = (u - {(x - y)}^{2})

C O N C L U S I O N

\sqrt{x} + \sqrt{y} = \sqrt{u}

i s e q u i v a l e n t a s o l v e

{4 x u = {(u + (x - y))}^{2}} \cup

{4 y u = {(u - (x - y))}^{2}}

Commented by manxsol last updated on 14/Dec/22

aplication (√(39−10a−a^2 ))+(√(9−a^2 ))=2(√(14)) x=39−10a−a^2 y=9−a^2 u=56 the magic : x−y=30−10a solve with 4yu=(u−(x−y))^2 (it easy) 4(56)(9−a^2 )=[56−(30−10a)]^2 4(56)(9−a^2 )=[26+10a]^2 504−56a^2 =(13+5a)^2 504−56a^2 =169+25a^2 +130a 81a^2 +130a−335=0 a_1 = 1.3837 a_2 =−2.9887 root strange

a p l i c a t i o n

\sqrt{39 - 10 a - a^{2}} + \sqrt{9 - a^{2}} = 2 \sqrt{14}

x = 39 - 10 a - a^{2}

y = 9 - a^{2}

u = 56

t h e m a g i c : x - y = 30 - 10 a

s o l v e w i t h 4 y u = {(u - (x - y))}^{2}

(i t e a s y)

4 (56) (9 - a^{2}) = {[56 - (30 - 10 a)]}^{2}

4 (56) (9 - a^{2}) = {[26 + 10 a]}^{2}

504 - 56 a^{2} = {(13 + 5 a)}^{2}

504 - 56 a^{2} = 169 + 25 a^{2} + 130 a

81 a^{2} + 130 a - 335 = 0

a_{1} = 1.3837

a_{2} = - 2.9887 r o o t s t r a n g e

Answered by Frix last updated on 12/Dec/22

Generally: (√u)+(√v)=w Squaring and transforming ⇒ 2(√u)(√v)+v=w^2 −u−v Squaring and transforming ⇒ −4uv+(w^2 −u−v)^2 =0 Inserting u=−a^2 −10a+39∧v=−a^2 +9∧w=2(√(14)) 324a^2 +520a−1340=0 a^2 +((130a)/(81))−((335)/(81))=0 a>0 ⇒ a=((−65+56(√(10)))/(81))≈1.38379690 No “magic” possible

Generally : \sqrt{u} + \sqrt{v} = w

Squaring and transforming \Rightarrow

2 \sqrt{u} \sqrt{v} + v = w^{2} - u - v

Squaring and transforming \Rightarrow

- 4 u v + {(w^{2} - u - v)}^{2} = 0

Inserting u = - a^{2} - 10 a + 39 \land v = - a^{2} + 9 \land w = 2 \sqrt{14}

324 a^{2} + 520 a - 1340 = 0

a^{2} + \frac{130 a}{81} - \frac{335}{81} = 0

a > 0 \Rightarrow a = \frac{- 65 + 56 \sqrt{10}}{81} \approx 1.38379690

No “ {magic}^{″} possible

Commented by Acem last updated on 12/Dec/22

I thought in substituation the variable but i figured out that there will be still problems I ask for a magical way if is there?

I t h o u g h t i n s u b s t i t u a t i o n t h e v a r i a b l e b u t

i f i g u r e d o u t t h a t t h e r e w i l l b e s t i l l p r o b l e m s

I a s k f o r a m a g i c a l w a y i f i s t h e r e ?

Commented by Acem last updated on 12/Dec/22

Can you show all steps?

C a n y o u s h o w a l l s t e p s ?

Commented by Frix last updated on 12/Dec/22

I inserted a few lines more

I inserted a few lines more

Commented by Acem last updated on 13/Dec/22

Commented by Acem last updated on 13/Dec/22

How did you make the formula?

H o w d i d y o u m a k e t h e f o r m u l a ?

Answered by Acem last updated on 12/Dec/22

(√(39− 10 a −a^2 )) + (√(9−a^2 )) = 2 (√(14)) (√(10(3 −a) +(9−a^2 ))) + (√(9−a^2 )) = 2(√(14)) (√(9−a^2 )) ((√((10(3−a))/(9−a^2 ))) +1)= 2(√(14)) ((10(3−a))/(9−a^2 )) +1 + 2 (√((10(3−a))/(9−a^2 ))) = ((56)/(9−a^2 )) (√((10(3−a))/(9−a^2 ))) = (1/2)[ ((56)/(9−a^2 )) − ((10(3−a))/(9−a^2 )) −1]_(making it as one term) ^(No usefull with (3−a)(3+a)) 10 = (3+a)^ × S_2 ^( 2) ^↗ etc... etc until we get rid of the root To face another complicated problems 40= (((a^2 + 10a +17)^2 )/((3−a)^2 (3+a))) ∼ Boriiiing 40= (([(a+5)^2 −8]^2 )/((3−a)^2 (3+a))) Hfff

\sqrt{39 - 10 a - a^{2}} + \sqrt{9 - a^{2}} = 2 \sqrt{14}

\sqrt{10 (3 - a) + (9 - a^{2})} + \sqrt{9 - a^{2}} = 2 \sqrt{14}

\sqrt{9 - a^{2}} (\sqrt{\frac{10 (3 - a)}{9 - a^{2}}} + 1) = 2 \sqrt{14}

\frac{10 (3 - a)}{9 - a^{2}} + 1 + 2 \sqrt{\frac{10 (3 - a)}{9 - a^{2}}} = \frac{56}{9 - a^{2}}

\sqrt{\frac{10 (3 - a)}{9 - a^{2}}} = \underset{m a k i n g i t a s o n e t e r m}{\underset{⏟}{\frac{1}{2} [\frac{56}{9 - a^{2}} - \frac{10 (3 - a)}{9 - a^{2}} - 1]}}^{N o u s e f u l l w i t h (3 - a) (3 + a)}

10 = {(3 + a)}^{} \times S_{2}^{2}^{}

e t c \dots e t c u n t i l w e g e t r i d o f t h e r o o t

T o f a c e a n o t h e r c o m p l i c a t e d p r o b l e m s

40 = \frac{{(a^{2} + 10 a + 17)}^{2}}{{(3 - a)}^{2} (3 + a)} \sim B o r i i i i n g

40 = \frac{{[{(a + 5)}^{2} - 8]}^{2}}{{(3 - a)}^{2} (3 + a)} H f f f

Answered by Rasheed.Sindhi last updated on 13/Dec/22

(√(39−10a−a^2 )) +(√(9−a^2 )) =2(√(14)) _(−) ⇒∣a∣<3 ((√(39−10a−a^2 )) +(√(9−a^2 )) )((√(39−10a−a^2 )) −(√(9−a^2 )) ) =2(√(14)) ((√(39−10a−a^2 )) −(√(9−a^2 )) ) 39−10a−a^2 −9+a^2 =2(√(14)) ((√(39−10a−a^2 )) −(√(9−a^2 )) ) (√(39−10a−a^2 )) −(√(9−a^2 )) =((30−10a)/(2(√(14)))) (√(39−10a−a^2 )) −(√(9−a^2 )) =((15−5a)/( (√(14)))) determinant ((((√(39−10a−a^2 )) −(√(9−a^2 )) =((15−5a)/( (√(14))))_((√(39−10a−a^2 )) +(√(9−a^2 )) = ^( ) 2(√(14 )) ...(ii) ) ...(i)))) (ii)−(i) : 2(√(9−a^2 )) =2(√(14)) −((15−5a)/( (√(14)))) =((28−15+5a)/( (√(14))))=((13+5a)/( (√(14)))) {2((√(9−a^2 )))}^2 =(((13+5a)/( (√(14)))))^2 4(9−a^2 )=((169+130a+25a^2 )/(14)) 504−56a^2 =169+130a+25a^2 81a^2 +130a−335=0 a=((−130±(√(130^2 −4(81)(−335))))/(162)) a=((−65±56(√(10)))/(81)) a=1.3838^✓ ,−2.9888(Extraneous)

\underline{\sqrt{39 - 10 a - a^{2}} + \sqrt{9 - a^{2}} = 2 \sqrt{14}}

\Rightarrow∣ a ∣< 3

(\sqrt{39 - 10 a - a^{2}} + \sqrt{9 - a^{2}}) (\sqrt{39 - 10 a - a^{2}} - \sqrt{9 - a^{2}}) = 2 \sqrt{14} (\sqrt{39 - 10 a - a^{2}} - \sqrt{9 - a^{2}})

39 - 10 a - a^{2} - 9 + a^{2} = 2 \sqrt{14} (\sqrt{39 - 10 a - a^{2}} - \sqrt{9 - a^{2}})

\sqrt{39 - 10 a - a^{2}} - \sqrt{9 - a^{2}} = \frac{30 - 10 a}{2 \sqrt{14}}

\sqrt{39 - 10 a - a^{2}} - \sqrt{9 - a^{2}} = \frac{15 - 5 a}{\sqrt{14}}

\begin{matrix} \underset{\overset{}{\sqrt{39 - 10 a - a^{2}} + \sqrt{9 - a^{2}} =} 2 \sqrt{14} \dots (i i)}{\sqrt{39 - 10 a - a^{2}} - \sqrt{9 - a^{2}} = \frac{15 - 5 a}{\sqrt{14}}} \dots (i) \end{matrix}

(i i) - (i) : 2 \sqrt{9 - a^{2}} = 2 \sqrt{14} - \frac{15 - 5 a}{\sqrt{14}}

= \frac{28 - 15 + 5 a}{\sqrt{14}} = \frac{13 + 5 a}{\sqrt{14}}

{2 (\sqrt{9 - a^{2}})}^{2} = {(\frac{13 + 5 a}{\sqrt{14}})}^{2}

4 (9 - a^{2}) = \frac{169 + 130 a + 25 a^{2}}{14}

504 - 56 a^{2} = 169 + 130 a + 25 a^{2}

81 a^{2} + 130 a - 335 = 0

a = \frac{- 130 \pm \sqrt{130^{2} - 4 (81) (- 335)}}{162}

a = \frac{- 65 \pm 56 \sqrt{10}}{81}

a = 1 . 3838^{✓}, - 2.9888 (E x t r a n e o u s)

Commented by Acem last updated on 13/Dec/22

Weeeeee! Shokrannnn I will see the 2 your methods when finish work Thank you with all my heart!

W e e e e e e! S h o k r a n n n n

I w i l l s e e t h e 2 y o u r m e t h o d s w h e n f i n i s h w o r k

T h a n k y o u w i t h a l l m y h e a r t!

Commented by Rasheed.Sindhi last updated on 13/Dec/22

You′re welcome sir!

Y {ou}^{'} re w elcome s ir!

Answered by Rasheed.Sindhi last updated on 13/Dec/22

(√(39−10a−a^2 )) _(x) +(√(9−a^2 )) _(y) =2(√(14)) x+y=2(√(14)) ......(i) x^2 −y^2 =(39−10a−a^2 )−(9−a^2 )=30−10a x−y=((x^2 −y^2 )/(x+y))=((30−10a)/(2(√(14))))=((15−5a)/( (√(14))))...(ii) (i)−(ii): 2y=2(√(14)) −((15−5a)/( (√(14))))=((28−15+5a)/( (√(14)))) y=((13+5a)/(2(√(14)))) (√(9−a^2 )) =((13+5a)/(2(√(14)))) 9−a^2 =(((13+5a)/(2(√(14)))))^2 =((169+130a+25a^2 )/(56)) 504−56a^2 =169+130a+25a^2 81a^2 +130a−335=0 a=((−130±(√(130^2 −4(81)(−335))))/(162)) =((−130±112(√(10)) )/(162)) =((−65±56(√(10)) )/(81)) =1.3838^✓ , −2.9888(Not required)

\underset{x}{\underset{⏟}{\sqrt{39 - 10 a - a^{2}}}} + \underset{y}{\underset{⏟}{\sqrt{9 - a^{2}}}} = 2 \sqrt{14}

x + y = 2 \sqrt{14} \dots \dots (i)

x^{2} - y^{2} = (39 - 10 a - a^{2}) - (9 - a^{2}) = 30 - 10 a

x - y = \frac{x^{2} - y^{2}}{x + y} = \frac{30 - 10 a}{2 \sqrt{14}} = \frac{15 - 5 a}{\sqrt{14}} \dots (i i)

(i) - (i i) :

2 y = 2 \sqrt{14} - \frac{15 - 5 a}{\sqrt{14}} = \frac{28 - 15 + 5 a}{\sqrt{14}}

y = \frac{13 + 5 a}{2 \sqrt{14}}

\sqrt{9 - a^{2}} = \frac{13 + 5 a}{2 \sqrt{14}}

9 - a^{2} = {(\frac{13 + 5 a}{2 \sqrt{14}})}^{2} = \frac{169 + 130 a + 25 a^{2}}{56}

504 - 56 a^{2} = 169 + 130 a + 25 a^{2}

81 a^{2} + 130 a - 335 = 0

a = \frac{- 130 \pm \sqrt{130^{2} - 4 (81) (- 335)}}{162}

= \frac{- 130 \pm 112 \sqrt{10}}{162}

= \frac{- 65 \pm 56 \sqrt{10}}{81}

= 1 . 3838^{✓}, - 2.9888 (N o t r e q u i r e d)

Commented by Acem last updated on 13/Dec/22

Thank you! dear friend

T h a n k y o u! d e a r f r i e n d

Commented by manxsol last updated on 13/Dec/22

wait,Sir Acem other method

w a i t, S i r A c e m o t h e r m e t h o d

Commented by manxsol last updated on 13/Dec/22

solution not is completed

s o l u t i o n n o t i s c o m p l e t e d

Commented by manxsol last updated on 13/Dec/22

I found the magic against boredom

I f o u n d t h e m a g i c a g a i n s t b o r e d o m

Answered by mr W last updated on 14/Dec/22

Commented by mr W last updated on 13/Dec/22

geometric method (√(39−10x−x^2 ))+(√(9−x^2 ))=2(√(14)) ⇒(√(8^2 −(5+x)^2 ))+(√(3^2 −x^2 ))=2(√(14)) the geometric meaning of this equation see diagram above. (√(5^2 +(2(√(14)))^2 ))=9 sin α=(5/9) ⇒cos α=((2(√(14)))/9) cos β=((8^2 +9^2 −3^2 )/(2×8×9))=((17)/(18)) ⇒sin β=((√(35))/(18)) 5+x=8 sin (α+β) =8 ((5/9)×((17)/(18))+((2(√(14)))/9)×((√(35))/(18))) =((4(85+14(√(10))))/(81)) ⇒x=((4(85+14(√(10))))/(81))−5=((56(√(10))−65)/(81)) ✓

\underset{―}{g e o m e t r i c m e t h o d}

\sqrt{39 - 10 x - x^{2}} + \sqrt{9 - x^{2}} = 2 \sqrt{14}

\Rightarrow \sqrt{8^{2} - {(5 + x)}^{2}} + \sqrt{3^{2} - x^{2}} = 2 \sqrt{14}

t h e g e o m e t r i c m e a n i n g o f t h i s

e q u a t i o n s e e d i a g r a m a b o v e .

\sqrt{5^{2} + {(2 \sqrt{14})}^{2}} = 9

\sin α = \frac{5}{9} \Rightarrow \cos α = \frac{2 \sqrt{14}}{9}

\cos β = \frac{8^{2} + 9^{2} - 3^{2}}{2 \times 8 \times 9} = \frac{17}{18} \Rightarrow \sin β = \frac{\sqrt{35}}{18}

5 + x = 8 \sin (α + β)

= 8 (\frac{5}{9} \times \frac{17}{18} + \frac{2 \sqrt{14}}{9} \times \frac{\sqrt{35}}{18})

= \frac{4 (85 + 14 \sqrt{10})}{81}

\Rightarrow x = \frac{4 (85 + 14 \sqrt{10})}{81} - 5 = \frac{56 \sqrt{10} - 65}{81} ✓

Commented by manxsol last updated on 14/Dec/22

you are great!

y o u a r e g r e a t!

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