To avoid reducing to the same denominator, we have just to replace 1/(x+2)^2 by x^2-8x+16, because 1/(x+2) =4-x and squaring both sides gives 1/(x+2)^2 = x^2-8x+16. Then, x^2+4x+1/(x+2)^2=x^2+4x+x^2-8x+16 =2x^2-4x+16. We replace x^2 by 2x+7 and then we get 2(2x+7)-4x+16=4x+14-4x+16=30.
I used the substitution and it worked Let x +2 =u => x = u-2 So our first equation becomes U-2 +1/u =4 On simplifying we get U+1/U =6 Square both sides We get u⁴+1=34 u²-------(1) Now , because of substitution second equation on simplifying becomes u⁴+1 -4u²/u² We know the value of u⁴+1 from eq 1 So we get 34u²-4u²/u² =30 😊😊😊
Solving the first eq. gives x = 1 ± √8. We rearrange this eq. then to get 1 / (x + 2)² = (4 - x)² = (3 ∓ √8)² = 17 ∓ 6√8. We now use both results for the second eq. and get (1 ± √8)² + 4 (1 ± √8) + 17 ∓ 6√8 = 9 ± 2√8 + 4 ± 4√8 + 17 ∓ 6√8 = 30.
using the substitution x=u-2 the two equations become u-2+1/u=4 -> u+1/u=6 and (u-2)^2+4(u-2)+1/u^2=? u^2-4u+4+4u-8+1/u^2=? u^2+1/u^2-4=? squaring the first equation we get u^2+2+1/u^2=36 or u^2+1/u^2=34 thus we get u^2+1/u^2-4=34-4=30
To avoid reducing to the same denominator, we have just to replace 1/(x+2)^2 by x^2-8x+16, because 1/(x+2) =4-x and squaring both sides gives 1/(x+2)^2 = x^2-8x+16. Then, x^2+4x+1/(x+2)^2=x^2+4x+x^2-8x+16 =2x^2-4x+16. We replace x^2 by 2x+7 and then we get 2(2x+7)-4x+16=4x+14-4x+16=30.
I used the substitution and it worked
Let x +2 =u
=> x = u-2
So our first equation becomes
U-2 +1/u =4
On simplifying we get
U+1/U =6
Square both sides
We get u⁴+1=34 u²-------(1)
Now , because of substitution second equation on simplifying becomes
u⁴+1 -4u²/u²
We know the value of u⁴+1 from eq 1
So we get 34u²-4u²/u² =30 😊😊😊
Solving the first eq. gives x = 1 ± √8. We rearrange this eq. then to get 1 / (x + 2)² = (4 - x)² = (3 ∓ √8)² = 17 ∓ 6√8. We now use both results for the second eq. and get (1 ± √8)² + 4 (1 ± √8) + 17 ∓ 6√8 = 9 ± 2√8 + 4 ± 4√8 + 17 ∓ 6√8 = 30.
using the substitution x=u-2 the two equations become
u-2+1/u=4 -> u+1/u=6
and
(u-2)^2+4(u-2)+1/u^2=?
u^2-4u+4+4u-8+1/u^2=?
u^2+1/u^2-4=?
squaring the first equation we get
u^2+2+1/u^2=36 or u^2+1/u^2=34
thus we get
u^2+1/u^2-4=34-4=30
x + 1/(x + 2) = 4
find
x² + 4x + 1/(x + 2)²
(x + 2) + 1/(x + 2) = 6
x² + 4x + 1/(x + 2)² = A
A = (x + 2)² + 1/(x + 2)² - 4
(x + 2) + 1/(x + 2) = 6
(x + 2)² + 1/(x + 2)² + 2 = 36
(x + 2)² + 1/(x + 2)² = 34
A = (x + 2)² + 1/(x + 2)² - 4
A = 34 - 4
*A = 30*
Aw, you're throwing us a softball. u = (x+2) + 1/(x+2) etc etc😊
Why not solve for x? It’s pretty easy.
3η Μεθοδος
X+2=Y , X=Y-2, Y-2+1/Y=4
Y+1/Y=6, (Y+1/Y)^2=6^2
Y^2+2+(1/Y)^2=36, Y^2+(1/Y)^2=34
(Y-2)^2+4(Y-2)+(1/Y)^2=
Y^2-4Y+4+4Y-8+(1/Y)^2=
Y^2+(1/Y)^2-4=34-4=30
😊😊😊👍👍👍