To solve the equation [tex]\(x \cdot |2x + 3| = 5\)[/tex], we need to break it down into different cases based on the nature of the absolute value function. The absolute value function splits into two cases depending on whether the expression inside it is non-negative or negative.
### Case 1: [tex]\(2x + 3 \geq 0\)[/tex]
If [tex]\(2x + 3 \geq 0\)[/tex], then [tex]\(|2x + 3| = 2x + 3\)[/tex]. The equation becomes:
[tex]\[ x (2x + 3) = 5 \][/tex]
Expanding and rearranging terms:
[tex]\[ 2x^2 + 3x - 5 = 0 \][/tex]
This is a quadratic equation of the form [tex]\(ax^2 + bx + c = 0\)[/tex]. To solve this quadratic equation, we can use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substituting [tex]\(a = 2\)[/tex], [tex]\(b = 3\)[/tex], and [tex]\(c = -5\)[/tex]:
[tex]\[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-5)}}{2 \cdot 2} \][/tex]
[tex]\[ x = \frac{-3 \pm \sqrt{9 + 40}}{4} \][/tex]
[tex]\[ x = \frac{-3 \pm \sqrt{49}}{4} \][/tex]
[tex]\[ x = \frac{-3 \pm 7}{4} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{-3 + 7}{4} = 1 \][/tex]
[tex]\[ x = \frac{-3 - 7}{4} = -2.5 \][/tex]
We need to check which solutions satisfy the original condition [tex]\(2x + 3 \geq 0\)[/tex]:
- For [tex]\(x = 1\)[/tex]: [tex]\(2(1) + 3 = 5 \geq 0\)[/tex], so [tex]\(x = 1\)[/tex] is valid.
- For [tex]\(x = -2.5\)[/tex]: [tex]\(2(-2.5) + 3 = -2\)[/tex], which does not satisfy [tex]\(2x + 3 \geq 0\)[/tex].
So, the valid solution from this case is [tex]\(x = 1\)[/tex].
### Case 2: [tex]\(2x + 3 < 0\)[/tex]
If [tex]\(2x + 3 < 0\)[/tex], then [tex]\(|2x + 3| = -(2x + 3)\)[/tex]. The equation becomes:
[tex]\[ x (-(2x + 3)) = 5 \][/tex]
Simplifying:
[tex]\[ -2x^2 - 3x = 5 \][/tex]
[tex]\[ 2x^2 + 3x + 5 = 0 \][/tex]
To solve for [tex]\(x\)[/tex], we again use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\(a = 2\)[/tex], [tex]\(b = 3\)[/tex], and [tex]\(c = 5\)[/tex]:
[tex]\[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot 5}}{2 \cdot 2} \][/tex]
[tex]\[ x = \frac{-3 \pm \sqrt{9 - 40}}{4} \][/tex]
[tex]\[ x = \frac{-3 \pm \sqrt{-31}}{4} \][/tex]
Since the discriminant [tex]\((9 - 40 = -31)\)[/tex] is negative, there are no real solutions for this case.
### Conclusion
From the above analysis, the only real solutions to the equation [tex]\(x \cdot |2x + 3| = 5\)[/tex] are:
[tex]\[ x = -\frac{5}{2} \quad \text{and} \quad x = 1 \][/tex]
Therefore, the final solutions are:
[tex]\[ x = -\frac{5}{2} \quad \text{and} \quad x = 1 \][/tex]
So, among the given choices:
- [tex]\(x = -4, x = 1\)[/tex]
- [tex]\(x = 1\)[/tex]
- [tex]\(x = -1, x = 4\)[/tex]
- No solution
We can see that the correct choice is [tex]\(x=1\)[/tex].
