For-what-values-of-a-does-the-system-of-equations-only-have-one-solution-a-x-4-1-y-2-x-x-2-y-2-4-

Question Number 181182 by depressiveshrek last updated on 22/Nov/22

F o r w h a t v a l u e s o f a d o e s t h e s y s t e m

o f e q u a t i o n s o n l y h a v e o n e s o l u t i o n :

{\begin{cases} a (x^{4} + 1) = y + 2 - ∣ x ∣ \\ x^{2} + y^{2} = 4 \end{cases}

Answered by mr W last updated on 22/Nov/22

y = a (x^{4} + 1) + ∣ x ∣ - 2

l e t t =∣ x ∣> 0

y = {a t}^{4} + t + a - 2 \dots (i)

y^{2} + t^{2} = 2^{2} \dots (i i)

(i) a n d (i i) s h o u l d t o u c h a t o n e a n d

o n l y o n e p o i n t . d u e t o s y m m e t r y,

t h i s p o i n t m u s t b e a t t = 0, i . e . (0, 2)

a (0^{4} + 1) + 0 - 2 = 2

\Rightarrow a = 4 ✓

Commented by mr W last updated on 22/Nov/22

Commented by depressiveshrek last updated on 22/Nov/22

T h a n k y o u s i r, y o u a r e h e l p i n g m e

a l o t . C a n I c o n t a c t y o u s o m e h o w ?

I n e e d h e l p t o p r a c t i c e f o r m y

{c o u n t r y}^{'} s n a t i o n a l m a t h o l y m p i a d .

Commented by mr W last updated on 22/Nov/22

h e r e i s a g o o d p l a c e . a l l p e o p l e c a n

h e l p y o u .

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