Answer :
To find the real solutions of the quadratic equation [tex]\(2x^2 + 2x - 2 = 0\)[/tex], we follow these steps:
1. Identify the coefficients of the quadratic equation:
[tex]\[ a = 2, \quad b = 2, \quad c = -2 \][/tex]
2. Calculate the discriminant:
The discriminant 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]\[ \Delta = 2^2 - 4 \cdot 2 \cdot (-2) = 4 + 16 = 20 \][/tex]
3. Determine the nature of the roots:
The discriminant tells us about the nature of the roots:
- If [tex]\(\Delta > 0\)[/tex], the equation has two distinct real solutions.
- If [tex]\(\Delta = 0\)[/tex], the equation has exactly one real solution (a repeated root).
- If [tex]\(\Delta < 0\)[/tex], the equation has no real solutions.
Since [tex]\(\Delta = 20 > 0\)[/tex], the equation [tex]\(2x^2 + 2x - 2 = 0\)[/tex] has two distinct real solutions.
4. Calculate the solutions using the quadratic formula:
The quadratic formula is:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
Substituting the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and the discriminant [tex]\(\Delta\)[/tex]:
[tex]\[ x_1 = \frac{-2 - \sqrt{20}}{2 \cdot 2} = \frac{-2 - \sqrt{20}}{4} \][/tex]
[tex]\[ x_2 = \frac{-2 + \sqrt{20}}{2 \cdot 2} = \frac{-2 + \sqrt{20}}{4} \][/tex]
5. Find the approximate numerical values of the solutions:
Simplifying the expressions:
[tex]\[ x_1 = \frac{-2 - \sqrt{20}}{4} = \frac{-2 - 2\sqrt{5}}{4} = \frac{-1 - \sqrt{5}}{2} \][/tex]
[tex]\[ x_2 = \frac{-2 + \sqrt{20}}{4} = \frac{-2 + 2\sqrt{5}}{4} = \frac{-1 + \sqrt{5}}{2} \][/tex]
Evaluating these expressions:
[tex]\[ x_1 \approx -1.618033988749895 \][/tex]
[tex]\[ x_2 \approx 0.6180339887498949 \][/tex]
6. Conclusion:
The quadratic equation [tex]\(2x^2 + 2x - 2 = 0\)[/tex] has real solutions.
[tex]\[ \text{Yes} \][/tex]
[tex]\[ \text{Solutions: } x_1 = -1.618033988749895, \quad x_2 = 0.6180339887498949 \][/tex]
So, the answer to the given problem is:
- The equation has real solutions: Yes.
- The solutions are [tex]\(x_1 = -1.618033988749895\)[/tex] and [tex]\(x_2 = 0.6180339887498949\)[/tex] with [tex]\(x_1 \leq x_2\)[/tex].
1. Identify the coefficients of the quadratic equation:
[tex]\[ a = 2, \quad b = 2, \quad c = -2 \][/tex]
2. Calculate the discriminant:
The discriminant 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]\[ \Delta = 2^2 - 4 \cdot 2 \cdot (-2) = 4 + 16 = 20 \][/tex]
3. Determine the nature of the roots:
The discriminant tells us about the nature of the roots:
- If [tex]\(\Delta > 0\)[/tex], the equation has two distinct real solutions.
- If [tex]\(\Delta = 0\)[/tex], the equation has exactly one real solution (a repeated root).
- If [tex]\(\Delta < 0\)[/tex], the equation has no real solutions.
Since [tex]\(\Delta = 20 > 0\)[/tex], the equation [tex]\(2x^2 + 2x - 2 = 0\)[/tex] has two distinct real solutions.
4. Calculate the solutions using the quadratic formula:
The quadratic formula is:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
Substituting the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and the discriminant [tex]\(\Delta\)[/tex]:
[tex]\[ x_1 = \frac{-2 - \sqrt{20}}{2 \cdot 2} = \frac{-2 - \sqrt{20}}{4} \][/tex]
[tex]\[ x_2 = \frac{-2 + \sqrt{20}}{2 \cdot 2} = \frac{-2 + \sqrt{20}}{4} \][/tex]
5. Find the approximate numerical values of the solutions:
Simplifying the expressions:
[tex]\[ x_1 = \frac{-2 - \sqrt{20}}{4} = \frac{-2 - 2\sqrt{5}}{4} = \frac{-1 - \sqrt{5}}{2} \][/tex]
[tex]\[ x_2 = \frac{-2 + \sqrt{20}}{4} = \frac{-2 + 2\sqrt{5}}{4} = \frac{-1 + \sqrt{5}}{2} \][/tex]
Evaluating these expressions:
[tex]\[ x_1 \approx -1.618033988749895 \][/tex]
[tex]\[ x_2 \approx 0.6180339887498949 \][/tex]
6. Conclusion:
The quadratic equation [tex]\(2x^2 + 2x - 2 = 0\)[/tex] has real solutions.
[tex]\[ \text{Yes} \][/tex]
[tex]\[ \text{Solutions: } x_1 = -1.618033988749895, \quad x_2 = 0.6180339887498949 \][/tex]
So, the answer to the given problem is:
- The equation has real solutions: Yes.
- The solutions are [tex]\(x_1 = -1.618033988749895\)[/tex] and [tex]\(x_2 = 0.6180339887498949\)[/tex] with [tex]\(x_1 \leq x_2\)[/tex].
