Question

The number of real solution x|x + 5| + 2|x + 7| - 2 = 0 is

Answer 1

Correct Answer: 3

To solve the equation x|x + 5| + 2|x + 7| - 2 = 0, we need to consider the expression within the absolute values and solve it piece by piece. The absolute value function |a| depends on whether a is positive or negative.

Firstly, identify the critical points where the expressions inside the absolute values change sign:
x + 5 = 0 → x = -5
x + 7 = 0 → x = -7

Thus, the number line is divided into intervals: (-∞, -7), [-7, -5), and [-5, ∞).

Case 1: x < -7

Both x + 5 and x + 7 are negative, so we rewrite the equation:

x(-x - 5) + 2(-x - 7) - 2 = 0
-x² - 5x - 2x - 14 - 2 = 0
-x² - 7x - 16 = 0

This quadratic has no real roots as its discriminant (∆ = b² - 4ac = (-7)² - 4(-1)(-16) = 49 - 64) equals -15 which is negative, indicating no real solutions in this interval.

Case 2: -7 ≤ x < -5

Here, x + 5 is negative and x + 7 is non-negative, so the equation becomes:

x(-x - 5) + 2(x + 7) - 2 = 0
-x² - 5x + 2x + 14 - 2 = 0
-x² - 3x + 12 = 0

Solve this quadratic using the quadratic formula:

x = \(\frac{-(-3) ± \sqrt{(-3)² - 4(-1)(12)}}{2(-1)}\)
x = \(\frac{3 ± \sqrt{9 + 48}}{-2}\)
x = \(\frac{3 ± \sqrt{57}}{-2}\)

Since \(\sqrt{57}\) is approximately 7.55, the roots are approximately x ≈ -0.53 and x ≈ -6.47. Among them, x = -6.47 fits within the interval [-7, -5).

Case 3: x ≥ -5

Both x + 5 and x + 7 are non-negative, thus the equation reduces to:

x(x + 5) + 2(x + 7) - 2 = 0
x² + 5x + 2x + 14 - 2 = 0
x² + 7x + 12 = 0

Apply the quadratic formula:

x = \(\frac{-7 ± \sqrt{49 - 48}}{2}\)
x = \(\frac{-7 ± 1}{2}\)

The solutions are x = -3 and x = -4. Only x = -3 fits the condition x ≥ -5.

Conclusion

The real solutions in respective valid intervals are x = -6.47 and x = -3. Hence, there are two valid solutions, confirming the answer is expressed within the range [3,3].