Question

При каком наименьшем с уравнение х^2-4х+2с+1=0 не имеет действительных корней? Ответы:1.1;2.2;3.3;4.4

Answer 1

Уравнение не имеет корней, если дискриминант меньше 0.
$16-4(2c+1)\ \textless \ 0|:-4\\2c+1\ \textgreater \ 4\\2c\ \textgreater \ 3\\c\ \textgreater \ 1,5=\ \textgreater \ c=2$

Answer 2

Уравнение не имеет действительных корней при условии если дискриминант меньше 0
D=(-4)²-4*1*(2c+1)=16-8c-4=12-8c
12-8c<0
-8c<-12
c>1,5
c∈(1,5;∞)
Ответ наименьшее значение с=2

Answer 3

Finding the Minimum Value for Real Roots

To find the minimum value for which the equation \(x^2 - 4x + 2c + 1 = 0\) does not have real roots, we can use the discriminant of the quadratic equation. The discriminant is given by the formula:

\[ D = b^2 - 4ac \]

Where \(a = 1\), \(b = -4\), and \(c = 2c + 1\). For the equation to have no real roots, the discriminant must be negative.

Calculating the Discriminant

Let's calculate the discriminant using the given equation:

\[ D = (-4)^2 - 4*1*(2c + 1) \]

Simplifying this expression gives:

\[ D = 16 - 8c - 4 \] \[ D = 12 - 8c \]

Condition for No Real Roots

For the equation to have no real roots, the discriminant \(D\) must be less than zero. Therefore, we have the inequality:

\[ 12 - 8c < 0 \]

Solving for \(c\), we get:

\[ -8c < -12 \] \[ c > \frac{12}{8} \] \[ c > \frac{3}{2} \]

Conclusion

Therefore, the equation \(x^2 - 4x + 2c + 1 = 0\) does not have real roots when \(c > \frac{3}{2}\).

Answer: The equation does not have real roots for \(c > \frac{3}{2}\).