Πρεπει χ=/-2. Θετω χ^2+4χ-20=ψ. Η εξισωση γραφεται:[ψ-3(χ+2)]^2/(χ+2)^2+[ψ+3(χ+2)]^2/(χ+2)^2=36 θετωα=[ψ/(χ+2)]-3 και β=[ψ/(χ+2)]+3 εχω το συστημα α-β= -6 και α^2+β^2=36. Που δινει α=0 ; β=6 ή α=-6; β=0. Οποτε τελικα χ=[-1+ -(105)^(1/2)]/2 ή χ=[-7+ -(105)^(1/2)]/2.
{[(x² +4x-20)-3(x+2)]/(x+2)}² +{[(x²+4x-20)+3(x+2)]/(x+2)}² = 36 or [(x²+4x-20)/(x+2)+3]²+[(x²+4x-20)/(x+2)-3]² = 36 (1) with x≠ -2 . Let (x²+4x-20)/(x+2) = t and the (1) is equivalent to (t+3)² + (t-3)² = 36 or t² = 9 or t = ±3 after simplifications. So (x²+4x-20)/(x+2) = ± 3 . The (x² +4x-20)(x+2) = 3 => x² +4x -20=3(x+2) => x² +x-26=0 => x = (-1±√105)/2 . The (x²+4x-20)/(x+2)=-3 => x²+4x-20=-3(x+2) => x² +7x-14 =0 => x = (-7±√105)/2 . Both of them roots are accepted. Remark. Why x²+4x-20 ... Trick [(x²+x-26)+(x²+7x-14)]/2 = x²+4x-20 etc ..
Let 1st n 2nd term is a &b.then
a^2+b^2= 36 & a-b= -6 solving gives a= 0; - 6 & b= 6; 0 gives x= (-1+ -√105)/2 ;
X= (-7+ - √105 )/2 solns.
❤ outstanding observation
Very nice
Πρεπει χ=/-2. Θετω χ^2+4χ-20=ψ. Η εξισωση γραφεται:[ψ-3(χ+2)]^2/(χ+2)^2+[ψ+3(χ+2)]^2/(χ+2)^2=36 θετωα=[ψ/(χ+2)]-3 και β=[ψ/(χ+2)]+3 εχω το συστημα α-β= -6 και α^2+β^2=36. Που δινει α=0 ; β=6 ή α=-6; β=0. Οποτε τελικα χ=[-1+ -(105)^(1/2)]/2 ή χ=[-7+ -(105)^(1/2)]/2.
X=[-1+(105)^(1/2)]/2, -1-(105)^(1/2)]/2, [-7+(105)^(1/2)]/2, [-7-(105)^(1/2)]/2.
Shorts.. ..ab = 0 where a& b r the terms.of eqn.
Hence both r the solns.
at 1.06 was that a toilet flush i heard?
Con bò
[(x² + x - 26)/(x + 2)]² + [(x² + 7x - 14)/(x + 2)]² = 36 → wher
[(x² + x - 26)² + (x² + 7x - 14)²]/(x + 2)² = 36
(x² + x - 26)² + (x² + 7x - 14)² = 36.(x + 2)²
(x² + x - 26)² + (x² + 7x - 14)² - 36.(x + 2)² = 0
(x² + x - 26)² + (x² + 7x - 14)² - 6².(x + 2)² = 0
(x² + x - 26)² + [(x² + 7x - 14) + 6.(x + 2)].[(x² + 7x - 14) - 6.(x + 2)] = 0
(x² + x - 26)² + [x² + 7x - 14 + 6x + 12].[x² + 7x - 14 - 6x - 12] = 0
(x² + x - 26)² + (x² + 13x - 2).(x² + x - 26) = 0
(x² + x - 26).[(x² + x - 26) + (x² + 13x - 2)] = 0
(x² + x - 26).(x² + x - 26 + x² + 13x - 2) = 0
(x² + x - 26).[2x² + 14x - 28] = 0
(x² + x - 26) * 2.(x² + 7x - 14) = 0
(x² + x - 26).(x² + 7x - 14) = 0
First case: (x² + x - 26) = 0
x² + x - 26 = 0
Δ = (1)² - (4 * - 26) = 105
x = (- 1 ± √105)/2
Second case: (x² + 7x - 14) = 0
x² + 7x - 14 = 0
Δ = (7)² - (4 * - 14) = 105
x = (- 7 ± √105)/2
{[(x² +4x-20)-3(x+2)]/(x+2)}² +{[(x²+4x-20)+3(x+2)]/(x+2)}² = 36 or
[(x²+4x-20)/(x+2)+3]²+[(x²+4x-20)/(x+2)-3]² = 36 (1) with x≠ -2 .
Let (x²+4x-20)/(x+2) = t and the (1) is equivalent to (t+3)² + (t-3)² = 36 or
t² = 9 or t = ±3 after simplifications.
So (x²+4x-20)/(x+2) = ± 3 .
The (x² +4x-20)(x+2) = 3 =>
x² +4x -20=3(x+2) => x² +x-26=0 =>
x = (-1±√105)/2 .
The (x²+4x-20)/(x+2)=-3 =>
x²+4x-20=-3(x+2) =>
x² +7x-14 =0 =>
x = (-7±√105)/2 .
Both of them roots are accepted.
Remark.
Why x²+4x-20 ...
Trick
[(x²+x-26)+(x²+7x-14)]/2 = x²+4x-20 etc ..