Question

2x^2+5x-3>0 x^2 >=64

Answer 1

решение задания смотри на фотографии

Answer 2

To solve the inequalities $2x^2 + 5x - 3 > 0$ and $x^2 \geq 64$, we'll break them down step by step.

1. Solving $2x^2 + 5x - 3 > 0$:

To solve this quadratic inequality, we need to find the intervals where $2x^2 + 5x - 3$ is greater than zero.

First, let's find the roots of the quadratic equation $2x^2 + 5x - 3 = 0$. The roots can be found using the quadratic formula:

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

where $a = 2$, $b = 5$, and $c = -3$:

$x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2}$ $x = \frac{-5 \pm \sqrt{25 + 24}}{4}$ $x = \frac{-5 \pm \sqrt{49}}{4}$ $x = \frac{-5 \pm 7}{4}$

So, the roots are:

$x_1 = \frac{-5 + 7}{4} = \frac{2}{4} = \frac{1}{2}$ $x_2 = \frac{-5 - 7}{4} = \frac{-12}{4} = -3$

Now, the inequality $2x^2 + 5x - 3 > 0$ can be expressed as:

$(x - \frac{1}{2})(x + 3) > 0$

Next, we use the concept of sign charts to find the intervals where the expression is greater than zero:

$$\begin{array}{c|cccc} x & -\infty & \frac{1}{2} & -3 & \infty \\ \hline (x - \frac{1}{2}) & - & 0 & - & - \\ (x + 3) & - & + & + & + \\ \hline \text{Expression} & + & 0 & - & + \end{array}$$

From the sign chart, we see that the expression is positive in two intervals: $\frac{1}{2} < x < -3$ and $x > \frac{1}{2}$.

Therefore, the solution to the inequality $2x^2 + 5x - 3 > 0$ is $x \in \left(\frac{1}{2}, \infty\right)$.

2. Solving $x^2 \geq 64$:

To solve this inequality, we'll move all terms to one side to obtain:

$x^2 - 64 \geq 0$

Now, we factor the left side:

$x^2 - 64 = (x + 8)(x - 8) \geq 0$

Next, we use the concept of sign charts again to find the intervals where the expression is greater than or equal to zero:

$$\begin{array}{ccccc}x& -\mathrm{\infty }& -8& 8& \mathrm{\infty }\end{array}$$