Question

2 икс в квадрате минус 5 икс плюс 3 равно 0 : икс в квадрате минус 5 минус 1 равно 0 : икс в квадрате плюс 3 икс плюс 1 равно 0: 3икс в квадрате плюс 7 икс минус 6 равно 0 : 2икс в квадрат е -9икс плюс 4 равно 0 . Кто ответит на все отдам все баллы зарание спасибо

Answer 1

1)
2x²-5x+3=0
Так как сумма коэффициентов равна нулю, то
x=1, x=3/2

2)
x²-5-1=0
x²=6
x=√6, x=-√6

3)
x²+3x+1=0
D=3²-4*1*1=5
x=(-3+√5)/2
x=(-3-√5)/2

4)
3x²+7x-6=0
D=7²-4*3*(-6)=49+72=121
x=(-7+11)/2*3=4/6=2/3
x=(-7-11)/2*3=-18/6=-3

5)
2x²-9x+4=0
D=9²-4*1*4=81-16=65
x=(9+√65)/4
x=(9-√65)/4

Answer 2

Quadratic Equations

To solve the given quadratic equations, let's analyze each equation separately:

1. 2x^2 - 5x + 3 = 0: To solve this equation, we can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a), where a, b, and c are the coefficients of the quadratic equation. In this case, a = 2, b = -5, and c = 3. Plugging in these values, we get: x = (-(-5) ± √((-5)^2 - 4 * 2 * 3)) / (2 * 2) Simplifying further: x = (5 ± √(25 - 24)) / 4 x = (5 ± √1) / 4 x = (5 ± 1) / 4 So, the solutions to this equation are: x = (5 + 1) / 4 = 6 / 4 = 1.5 x = (5 - 1) / 4 = 4 / 4 = 1

2. x^2 - 5 - 1 = 0: Simplifying this equation, we get: x^2 - 6 = 0 To solve this equation, we can add 6 to both sides: x^2 = 6 Taking the square root of both sides, we get: x = ±√6 So, the solutions to this equation are: x = √6 and x = -√6

3. x^2 + 3x + 1 = 0: To solve this equation, we can again use the quadratic formula. In this case, a = 1, b = 3, and c = 1. Plugging in these values, we get: x = (-3 ± √(3^2 - 4 * 1 * 1)) / (2 * 1) Simplifying further: x = (-3 ± √(9 - 4)) / 2 x = (-3 ± √5) / 2 So, the solutions to this equation are: x = (-3 + √5) / 2 and x = (-3 - √5) / 2

4. 3x^2 + 7x - 6 = 0: To solve this equation, we can again use the quadratic formula. In this case, a = 3, b = 7, and c = -6. Plugging in these values, we get: x = (-7 ± √(7^2 - 4 * 3 * -6)) / (2 * 3) Simplifying further: x = (-7 ± √(49 + 72)) / 6 x = (-7 ± √121) / 6 x = (-7 ± 11) / 6 So, the solutions to this equation are: x = (-7 + 11) / 6 = 4 / 6 = 2/3 x = (-7 - 11) / 6 = -18 / 6 = -3

5. 2x^2 - 9x + 4 = 0: To solve this equation, we can again use the quadratic formula. In this case, a = 2, b = -9, and c = 4. Plugging in these values, we get: x = (-(-9) ± √((-9)^2 - 4 * 2 * 4)) / (2 * 2) Simplifying further: x = (9 ± √(81 - 32)) / 4 x = (9 ± √49) / 4 x = (9 ± 7) / 4 So, the solutions to this equation are: x = (9 + 7) / 4 = 16 / 4 = 4 x = (9 - 7) / 4 = 2 / 4 = 0.5

Therefore, the solutions to the given quadratic equations are as follows: 1. 2x^2 - 5x + 3 = 0: x = 1.5, x = 1 2. x^2 - 5 - 1 = 0: x = √6, x = -√6 3. x^2 + 3x + 1 = 0: x = (-3 + √5) / 2, x = (-3 - √5) / 2 4. 3x^2 + 7x - 6 = 0: x = 2/3, x = -3 5. 2x^2 - 9x + 4 = 0: x = 4, x = 0.5

I hope this helps! Let me know if you have any further questions.