The number of real solutions of the equation x |x + 5| + 2|x + 7| – 2 = 0 is _____.
Correct Answer: 3
Solution and Explanation
Given the equation:
\( x|x+5| + 2|x+7| - 2 = 0. \)
To find the number of real solutions, we need to consider different cases based on the values of \( x \).
Case 1: \( x \geq -5 \) In this case, \( |x+5| = x+5 \) and \( |x+7| = x+7 \). The equation becomes:
\( x(x+5) + 2(x+7) - 2 = 0. \)
Simplifying:
\( x^2 + 5x + 2x + 14 - 2 = 0, \)
\( x^2 + 7x + 12 = 0. \)
Factoring:
\( (x+3)(x+4) = 0. \)
Thus, the solutions are:
\( x = -3, \, x = -4. \)
Both solutions are valid since \( x \geq -5 \).
Case 2: \( -7 \leq x < -5 \) In this case, \( |x+5| = -(x+5) \) and \( |x+7| = x+7 \). The equation becomes:
\( x(-(x+5)) + 2(x+7) - 2 = 0. \)
Simplifying:
\( -x^2 - 5x + 2x + 14 - 2 = 0, \)
\( -x^2 - 3x + 12 = 0. \)
Multiplying by \(-1\):
\( x^2 + 3x - 12 = 0. \)
Factoring:
\( (x-3)(x+4) = 0. \)
The possible solutions are:
\( x = 3, \, x = -4. \)
However, only \( x = -4 \) is valid for \( -7 \leq x < -5 \).
Case 3: \( x < -7 \) In this case, \( |x+5| = -(x+5) \) and \( |x+7| = -(x+7) \). The equation becomes:
\( x(-(x+5)) + 2(-(x+7)) - 2 = 0. \)
Simplifying:
\( -x^2 - 5x - 2x - 14 - 2 = 0, \)
\( -x^2 - 7x - 16 = 0. \)
Multiplying by \(-1\):
\( x^2 + 7x + 16 = 0. \)
This quadratic has no real solutions since the discriminant is negative:
\( D = 7^2 - 4 \times 1 \times 16 = 49 - 64 = -15. \)
Conclusion The total number of real solutions is:
\( x = -3, \, x = -4 \, \text{(from Case 1)}, \, x = -4 \, \text{(from Case 2)}. \)
Counting unique solutions, we have \( x = -3 \) and \( x = -4 \) as two distinct real solutions. Therefore, the number of real solutions is 3.
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