Question

The number of real solutions of the equation x |x + 5| + 2|x + 7| – 2 = 0 is _____.

Answer 1

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.