Question

1)2х² + 2х+3=0 2)х(х – 45) = 0 3)(х + 45)(х – 16) = 0 , 6) x2 – 51x = 0

Answer 1

Ответ:

2) x^2-45x=0. x^2=45x. x=0. x=45

Пошаговое объяснение:

Answer 2

It seems like you've provided a list of equations. It looks like you might want me to help you solve these equations for the variable x. I'll go through each equation one by one:

1. $2x^2 + 2x + 3 = 0$: This is a quadratic equation. To solve it, you can use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where the equation is in the form $ax^2 + bx + c = 0$. For this equation, $a = 2$, $b = 2$, and $c = 3$. Plugging in these values: $x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 2 \cdot 3}}{2 \cdot 2}$ $x = \frac{-2 \pm \sqrt{4 - 24}}{4}$ $x = \frac{-2 \pm \sqrt{-20}}{4}$ Since the discriminant ($b^2 - 4ac$) is negative, the solutions will be complex. The solutions are: $x = \frac{-1}{2} + \frac{\sqrt{5}}{2}i$ $x = \frac{-1}{2} - \frac{\sqrt{5}}{2}i$

2. $x(x - 45) = 0$: This equation is already factored. The solutions are the values of x that make each factor equal to zero: $x = 0$ $x - 45 = 0 \Rightarrow x = 45$

3. $(x + 45)(x - 16) = 0$: Similar to the previous equation, this one is already factored: $x + 45 = 0 \Rightarrow x = -45$ $x - 16 = 0 \Rightarrow x = 16$

4. $x^2 - 51x = 0$: This equation is also factored: $x(x - 51) = 0$ So, the solutions are: $x = 0$ $x - 51 = 0 \Rightarrow x = 51$

Feel free to ask if you have any more questions or if you need further clarification!