
3х + 2х^2– 5 = 0 х^2+ 2 + 3х = 0 х2+ 4х + 4 = 0 3х^2 + 8х = 3 6а^2+ 2 = 6а


Ответы на вопрос

Ответ:
1) Х1=-2,5; Х2=1
2) Х1=-2; Х2=-1
3) Х=-2
4) Х1=-3; Х2=1/3
5) Корней нет



Ответ:
-1 ; 2.5
нет корней
-2
1/3; -3
нет
Объяснение:
d=b²-4ac! x=-b±✓d/2a
d=9+4*2*5=49
x=-3±7/4=2.5 ;-1
и тд



Let's solve each of the given equations step by step:
- 3x + 2x^2 - 5 = 0
To solve this quadratic equation, we can rearrange it in the standard form:
2x^2 + 3x - 5 = 0
Now we can apply the quadratic formula to find the solutions for x:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 2, b = 3, and c = -5. Substituting these values into the quadratic formula, we have:
x = (-3 ± √(3^2 - 4 * 2 * -5)) / (2 * 2) x = (-3 ± √(9 + 40)) / 4 x = (-3 ± √49) / 4
Simplifying further:
x = (-3 ± 7) / 4
We get two solutions:
x1 = (-3 + 7) / 4 = 1 x2 = (-3 - 7) / 4 = -2
Therefore, the solutions to the equation 3x + 2x^2 - 5 = 0 are x = 1 and x = -2.
- x^2 + 2 + 3x = 0
We have a quadratic equation. Rearranging it:
x^2 + 3x + 2 = 0
We can factorize the quadratic equation:
(x + 2)(x + 1) = 0
Setting each factor equal to zero:
x + 2 = 0 --> x = -2 x + 1 = 0 --> x = -1
Therefore, the solutions to the equation x^2 + 2 + 3x = 0 are x = -2 and x = -1.
- x^2 + 4x + 4 = 0
This quadratic equation can be factored as a perfect square:
(x + 2)^2 = 0
Setting the square term equal to zero:
x + 2 = 0 --> x = -2
So the only solution to the equation x^2 + 4x + 4 = 0 is x = -2.
- 3x^2 + 8x = 3
Rearranging the equation:
3x^2 + 8x - 3 = 0
Using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 3, b = 8, and c = -3. Substituting these values into the formula:
x = (-8 ± √(8^2 - 4 * 3 * -3)) / (2 * 3) x = (-8 ± √(64 + 36)) / 6 x = (-8 ± √100) / 6
Simplifying further:
x = (-8 ± 10) / 6
We get two solutions:
x1 = (-8 + 10) / 6 = 1/3 x2 = (-8 - 10) / 6 = -3
Therefore, the solutions to the equation 3x^2 + 8x = 3 are x = 1/3 and x = -3.
- 6a^2 + 2 = 6a
Rearranging the equation:
6a^2 - 6a + 2 = 0
This quadratic equation cannot be easily factored or solved using the quadratic formula. We can use numerical methods such as the Newton-Raphson method or graphing techniques to approximate the solutions.


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