Answer:
x = 23
Explanation:
let's square both sides. that eliminates the individual square roots and creates one central term with square roots :
2x + 3 + x + 2 - 2 sqrt(2x + 3)×sqrt(x + 2) = 4
3x + 5 - 2×sqrt((2x + 3)(x + 2)) = 4
3x + 1 - 2×sqrt(2x² + 4x + 3x + 6) = 0
3x + 1 - 2×sqrt(2x² + 7x + 6) = 0
3x + 1 = 2×sqrt(2x² + 7x + 6)
now let's square both sides again
9x² + 6x + 1 = 4(2x² + 7x + 6) = 8x² + 28x + 24
x² - 22x - 23 = 0
a quadratic equation
ax²x+ bx + c = 0
has 2 solutions
x = (-b ± sqrt(b² - 4ac))/(2a)
in our case
x = (22 ± sqrt((-22)² - 4×1×-23))/(2×1) =
= (22 ± sqrt(484 + 92))/2 =
= (22 ± sqrt(576))/2 = (22 ± 24)/2 =
= 11 ± 12
x1 = 11 + 12 = 23
x2 = 11 - 12 = -1
since we squared twice in the process, and every solution to a square is a ±sqrt (so, 2 solutions with alternating signs), we need to try both solutions in the original equation to see which ones really fulfill it.
sqrt(2×23 + 3) - sqrt(23 + 2) = 2
sqrt(46 + 3) - sqrt(25) = 2
sqrt(49) - 5 = 2
7 - 5 = 2
correct, x = 23 is a valid solution.
sqrt(2×-1 + 3) - sqrt(-1 + 2) = 2
sqrt(-2 + 3) - sqrt(1) = 2
sqrt(1) - 1 = 2
1 - 1 = 2
wrong. so, x = -1 is not a valid solution to the original equation.