the number of real roots of the equation (x^2+2x)^2 -(x+1)^2-55 =0
Answers
Answered by
21
{ (x² +2x +1 ) -1} ² -(x +1)² -55 =0
(x +1)⁴ -2(x +1)² +1 -(x +1)² -55 =0
(x+1)⁴ -3(x +1)² -54 =0
let (x +1)²= r
r² -3r -54 =0
r² -9r +6r -54 =0
r( r -9) +6(r -9) =0
r = -6 , 9
but ( x+1)² ≥ 0 so, (x+1)² ≠ -6
so, (x +1)² = 9
x + 1 = ± 3
x = -1 ±3
x = -4, and 2
(x +1)⁴ -2(x +1)² +1 -(x +1)² -55 =0
(x+1)⁴ -3(x +1)² -54 =0
let (x +1)²= r
r² -3r -54 =0
r² -9r +6r -54 =0
r( r -9) +6(r -9) =0
r = -6 , 9
but ( x+1)² ≥ 0 so, (x+1)² ≠ -6
so, (x +1)² = 9
x + 1 = ± 3
x = -1 ±3
x = -4, and 2
Similar questions