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Sagot :
Sure, I can provide a detailed solution for the mathematical problem you've provided. The problem appears to be solving quadratic equations by factorization.
Firstly, you have an inequality and then a quadratic equation. Let's solve each part step-by-step.
### Part 1: Solve the inequality [tex]\( a^2 - 3a + 2 > 0 \)[/tex]
1. Factorize the quadratic expression: [tex]\( a^2 - 3a + 2 \)[/tex]
- To factorize, we need to find two numbers that multiply to the constant term (+2) and add up to the coefficient of the middle term (-3).
- The numbers that satisfy these conditions are -1 and -2.
Therefore, we can write:
[tex]\[ a^2 - 3a + 2 = (a - 1)(a - 2) \][/tex]
2. Set up the inequality:
[tex]\[ (a - 1)(a - 2) > 0 \][/tex]
3. Find the critical points: The inequality will change signs at the roots of the equation, which are [tex]\( a = 1 \)[/tex] and [tex]\( a = 2 \)[/tex].
4. Determine the intervals:
- The critical points divide the real number line into three intervals: [tex]\( (-\infty, 1) \)[/tex], [tex]\( (1, 2) \)[/tex], and [tex]\( (2, \infty) \)[/tex].
5. Test each interval:
- For [tex]\( a \in (-\infty, 1) \)[/tex]: Choose [tex]\( a = 0 \)[/tex]:
[tex]\[ (0 - 1)(0 - 2) = ( - 1)(-2) = 2 > 0 \][/tex]
Thus, the inequality holds in this interval.
- For [tex]\( a \in (1, 2) \)[/tex]: Choose [tex]\( a = 1.5 \)[/tex]:
[tex]\[ (1.5 - 1)(1.5 - 2) = 0.5 \times -0.5 = -0.25 < 0 \][/tex]
Thus, the inequality does not hold in this interval.
- For [tex]\( a \in (2, \infty) \)[/tex]: Choose [tex]\( a = 3 \)[/tex]:
[tex]\[ (3 - 1)(3 - 2) = 2 \times 1 = 2 > 0 \][/tex]
Thus, the inequality holds in this interval.
6. Conclusion:
[tex]\[ a^2 - 3a + 2 > 0 \implies a \in (-\infty, 1) \cup (2, \infty) \][/tex]
### Part 2: Solve the quadratic equation [tex]\(2e^2 - x + 10 = 0\)[/tex]
Assuming there is a typo in your expression and you meant to write it as [tex]\(2x^2 - x + 10 = 0\)[/tex]:
Given:
[tex]\[ 2x^2 - x + 10 = 0 \][/tex]
1. Use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 2 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = 10 \)[/tex].
2. Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac = (-1)^2 - 4(2)(10) = 1 - 80 = -79 \][/tex]
3. Since the discriminant is negative ([tex]\( \Delta = -79 \)[/tex]), the quadratic equation has no real solutions. It has complex solutions.
### Conclusion for the entire problem:
- For the inequality [tex]\( a^2 - 3a + 2 > 0 \)[/tex], the solution in the real number set is [tex]\( a \in (-\infty, 1) \cup (2, \infty) \)[/tex].
- For the quadratic equation [tex]\( 2x^2 - x + 10 = 0 \)[/tex], there are no real solutions, only complex solutions.
If you meant something else by [tex]\( 2e^2 - x + 10 = 0 \)[/tex], please provide clarification for a more accurate solution.
Firstly, you have an inequality and then a quadratic equation. Let's solve each part step-by-step.
### Part 1: Solve the inequality [tex]\( a^2 - 3a + 2 > 0 \)[/tex]
1. Factorize the quadratic expression: [tex]\( a^2 - 3a + 2 \)[/tex]
- To factorize, we need to find two numbers that multiply to the constant term (+2) and add up to the coefficient of the middle term (-3).
- The numbers that satisfy these conditions are -1 and -2.
Therefore, we can write:
[tex]\[ a^2 - 3a + 2 = (a - 1)(a - 2) \][/tex]
2. Set up the inequality:
[tex]\[ (a - 1)(a - 2) > 0 \][/tex]
3. Find the critical points: The inequality will change signs at the roots of the equation, which are [tex]\( a = 1 \)[/tex] and [tex]\( a = 2 \)[/tex].
4. Determine the intervals:
- The critical points divide the real number line into three intervals: [tex]\( (-\infty, 1) \)[/tex], [tex]\( (1, 2) \)[/tex], and [tex]\( (2, \infty) \)[/tex].
5. Test each interval:
- For [tex]\( a \in (-\infty, 1) \)[/tex]: Choose [tex]\( a = 0 \)[/tex]:
[tex]\[ (0 - 1)(0 - 2) = ( - 1)(-2) = 2 > 0 \][/tex]
Thus, the inequality holds in this interval.
- For [tex]\( a \in (1, 2) \)[/tex]: Choose [tex]\( a = 1.5 \)[/tex]:
[tex]\[ (1.5 - 1)(1.5 - 2) = 0.5 \times -0.5 = -0.25 < 0 \][/tex]
Thus, the inequality does not hold in this interval.
- For [tex]\( a \in (2, \infty) \)[/tex]: Choose [tex]\( a = 3 \)[/tex]:
[tex]\[ (3 - 1)(3 - 2) = 2 \times 1 = 2 > 0 \][/tex]
Thus, the inequality holds in this interval.
6. Conclusion:
[tex]\[ a^2 - 3a + 2 > 0 \implies a \in (-\infty, 1) \cup (2, \infty) \][/tex]
### Part 2: Solve the quadratic equation [tex]\(2e^2 - x + 10 = 0\)[/tex]
Assuming there is a typo in your expression and you meant to write it as [tex]\(2x^2 - x + 10 = 0\)[/tex]:
Given:
[tex]\[ 2x^2 - x + 10 = 0 \][/tex]
1. Use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 2 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = 10 \)[/tex].
2. Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac = (-1)^2 - 4(2)(10) = 1 - 80 = -79 \][/tex]
3. Since the discriminant is negative ([tex]\( \Delta = -79 \)[/tex]), the quadratic equation has no real solutions. It has complex solutions.
### Conclusion for the entire problem:
- For the inequality [tex]\( a^2 - 3a + 2 > 0 \)[/tex], the solution in the real number set is [tex]\( a \in (-\infty, 1) \cup (2, \infty) \)[/tex].
- For the quadratic equation [tex]\( 2x^2 - x + 10 = 0 \)[/tex], there are no real solutions, only complex solutions.
If you meant something else by [tex]\( 2e^2 - x + 10 = 0 \)[/tex], please provide clarification for a more accurate solution.
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