Question

Решите неравенство 5х²-8х+3>0

Answer 1

D=64-60=4

X1=(8+2):(2*5)=1

X2=(8-2):(2*5)=0,6

x>1

x>0,6

Ответ:

$(1 \: to \: + \infty )$

*to=;

Answer 2

To solve the inequality $5x^2 - 8x + 3 > 0$, we need to find the values of $x$ that satisfy the inequality. Here's how we can do it:

Step 1: Factorize the quadratic expression on the left-hand side: $5x^2 - 8x + 3 = (5x - 3)(x - 1)$

Step 2: Set each factor to zero and solve for $x$: $5x - 3 = 0 \Rightarrow x = \frac{3}{5}$ $x - 1 = 0 \Rightarrow x = 1$

Step 3: Now, plot these points on a number line:

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    3/5   1

Step 4: Pick a test point in each interval to determine the sign of the quadratic expression. For example, we can choose $x = 0$ (in the interval $-\infty, \frac{3}{5}$), $x = \frac{1}{2}$ (in the interval $\frac{3}{5}, 1$), and $x = 2$ (in the interval $1, \infty$).

Step 5: Substitute the test points into the original inequality and check the sign: For $x = 0$: $5(0)^2 - 8(0) + 3 = 3 > 0$ - True For $x = \frac{1}{2}$: $5\left(\frac{1}{2}\right)^2 - 8\left(\frac{1}{2}\right) + 3 = \frac{1}{4} - 4 + 3 = 0$ - False For $x = 2$: $5(2)^2 - 8(2) + 3 = 20 - 16 + 3 = 7 > 0$ - True

Step 6: Analyze the signs on the number line:

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    3/5   1

The inequality $5x^2 - 8x + 3 > 0$ is true in the intervals $-\infty, \frac{3}{5}$ and $1, \infty$.

Step 7: Write the final solution as an inequality: $x \in \left(-\infty, \frac{3}{5}\right) \cup (1, \infty)$

So, the solution to the inequality is $x \in \left(-\infty, \frac{3}{5}\right) \cup (1, \infty)$.