Find the number of real solutions of the equation _ x+1 (2x^2 + 7x + 5) + _ 2x+5 (x^2 + 2x + 1) = 4 .

Step-by-step solution

Given equation: _ x+1 (2x^2 + 7x + 5) + _ 2x+5 (x^2 + 2x + 1) = 4 First, factor the quadratic expressions in the arguments: 2x^2 + 7x + 5 = 2x^2 + 2x + 5x + 5 = 2x(x + 1) + 5(x + 1) = (2x + 5)(x + 1) x^2 + 2x + 1 = (x + 1)^2 Substitute these into the equation: _ x+1 ((2x + 5)(x + 1)) + _ 2x+5 ((x + 1)^2) = 4 Using properties of logarithms: _ x+1 (2x + 5) + _ x+1 (x + 1) + 2 _ 2x+5 (x + 1) = 4 _ x+1 (2x + 5) + 1 + 2 _ 2x+5 (x + 1) = 4 Let t = _ x+1 (2x + 5) . Then _ 2x+5 (x + 1) = 1 t . The equation becomes: t + 1 +