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Sagot :
To find the solutions of the quadratic equation [tex]\(2x^2 + 9x + 9 = 0\)[/tex], we can follow these steps:
1. Identify the coefficients: The quadratic equation is in the standard form [tex]\(ax^2 + bx + c = 0\)[/tex]. Here, the coefficients are:
- [tex]\(a = 2\)[/tex]
- [tex]\(b = 9\)[/tex]
- [tex]\(c = 9\)[/tex]
2. Calculate the discriminant: The discriminant [tex]\(\Delta\)[/tex] of a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[ \Delta = 9^2 - 4 \cdot 2 \cdot 9 \][/tex]
[tex]\[ \Delta = 81 - 72 \][/tex]
[tex]\[ \Delta = 9 \][/tex]
3. Determine the nature of the roots: The value of the discriminant indicates the nature of the roots of the quadratic equation.
- If [tex]\(\Delta > 0\)[/tex], there are two distinct real roots.
- If [tex]\(\Delta = 0\)[/tex], there is exactly one real root (a repeated root).
- If [tex]\(\Delta < 0\)[/tex], there are no real roots (the roots are complex).
Here, [tex]\(\Delta = 9\)[/tex], which means we have two distinct real roots.
4. Find the roots using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
Substitute [tex]\(a = 2\)[/tex], [tex]\(b = 9\)[/tex], and [tex]\(\Delta = 9\)[/tex]:
[tex]\[ x = \frac{-9 \pm \sqrt{9}}{2 \cdot 2} \][/tex]
Simplify under the square root:
[tex]\[ x = \frac{-9 \pm 3}{4} \][/tex]
This results in two solutions:
[tex]\[ x_1 = \frac{-9 + 3}{4} = \frac{-6}{4} = -1.5 \][/tex]
[tex]\[ x_2 = \frac{-9 - 3}{4} = \frac{-12}{4} = -3 \][/tex]
Therefore, the solutions to the quadratic equation [tex]\(2x^2 + 9x + 9 = 0\)[/tex] are:
[tex]\[ x = -1.5 \quad \text{and} \quad x = -3 \][/tex]
From the given choices, the correct answer is:
[tex]\(\boxed{x = -3, -\frac{3}{2}}\)[/tex]
1. Identify the coefficients: The quadratic equation is in the standard form [tex]\(ax^2 + bx + c = 0\)[/tex]. Here, the coefficients are:
- [tex]\(a = 2\)[/tex]
- [tex]\(b = 9\)[/tex]
- [tex]\(c = 9\)[/tex]
2. Calculate the discriminant: The discriminant [tex]\(\Delta\)[/tex] of a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[ \Delta = 9^2 - 4 \cdot 2 \cdot 9 \][/tex]
[tex]\[ \Delta = 81 - 72 \][/tex]
[tex]\[ \Delta = 9 \][/tex]
3. Determine the nature of the roots: The value of the discriminant indicates the nature of the roots of the quadratic equation.
- If [tex]\(\Delta > 0\)[/tex], there are two distinct real roots.
- If [tex]\(\Delta = 0\)[/tex], there is exactly one real root (a repeated root).
- If [tex]\(\Delta < 0\)[/tex], there are no real roots (the roots are complex).
Here, [tex]\(\Delta = 9\)[/tex], which means we have two distinct real roots.
4. Find the roots using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
Substitute [tex]\(a = 2\)[/tex], [tex]\(b = 9\)[/tex], and [tex]\(\Delta = 9\)[/tex]:
[tex]\[ x = \frac{-9 \pm \sqrt{9}}{2 \cdot 2} \][/tex]
Simplify under the square root:
[tex]\[ x = \frac{-9 \pm 3}{4} \][/tex]
This results in two solutions:
[tex]\[ x_1 = \frac{-9 + 3}{4} = \frac{-6}{4} = -1.5 \][/tex]
[tex]\[ x_2 = \frac{-9 - 3}{4} = \frac{-12}{4} = -3 \][/tex]
Therefore, the solutions to the quadratic equation [tex]\(2x^2 + 9x + 9 = 0\)[/tex] are:
[tex]\[ x = -1.5 \quad \text{and} \quad x = -3 \][/tex]
From the given choices, the correct answer is:
[tex]\(\boxed{x = -3, -\frac{3}{2}}\)[/tex]
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