Question

1)2х^2+3х+4=0 2)9х^2-6х+1=0

Answer 1

2x^2 + 3x + 4 = 0;
       D = b^2 - 4ac;
D = 3^2 - 4 * 2 * 4;
D = 9 - 32 = -23;
D < 0, корней нет!
       Ответ: x ∈ Ø.

9x^2 - 6x + 1 = 0;
       D = b^2 - 4ac;
D = -6^2 - 4 * 9 * 1;
D = 36 - 36 = 0;
D = 0, один корень!
       x = -b/2a;
x = 6/(2*9);
x = 6/18;  |:6
x = 1/3;
       Ответ: 1/3.

Answer 2

To solve these quadratic equations, you can use the quadratic formula, which is given by:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Where, in the general quadratic equation $ax^2 + bx + c = 0$:

• $a$ is the coefficient of $x^2$
• $b$ is the coefficient of $x$
• $c$ is the constant term

Let's apply this formula to your equations:

1. $2x^2 + 3x + 4 = 0$

In this equation, $a = 2$, $b = 3$, and $c = 4$.

$x = \frac{-3 \pm \sqrt{3^2 - 4(2)(4)}}{2(2)}$ $x = \frac{-3 \pm \sqrt{9 - 32}}{4}$ $x = \frac{-3 \pm \sqrt{-23}}{4}$

Since the discriminant ($b^2 - 4ac$) is negative, there are no real solutions. The solutions are complex. You can represent them in the form:

$x = \frac{-3 \pm i\sqrt{23}}{4}$

2. $9x^2 - 6x + 1 = 0$

In this equation, $a = 9$, $b = -6$, and $c = 1$.

$x = \frac{6 \pm \sqrt{(-6)^2 - 4(9)(1)}}{2(9)}$ $x = \frac{6 \pm \sqrt{36 - 36}}{18}$ $x = \frac{6 \pm \sqrt{0}}{18}$

Since the discriminant is zero, there is one real solution:

$x = \frac{6}{18} = \frac{1}{3}$

So, the solution to the second equation is $x = \frac{1}{3}$, and the first equation has complex solutions $x = \frac{-3 \pm i\sqrt{23}}{4}$.