Question

Решить неравенство:x^2 - 10x + 25 < 0 ​

Answer 1

Ответ:

${x}^{2} - 10x + 25 ={ (x - 5)}^{2} \\ { (x - 5)}^{2} < 0 \\ \\ x < 5 \\ x \: prinadlejit( - \infty .5)$

Answer 2

To solve the inequality $x^2 - 10x + 25 < 0$, we can use various methods. One way is to find the critical points of the quadratic function and then determine the intervals where the function is negative. Let's go through the steps:

Step 1: Factorize the quadratic expression, if possible. The given expression $x^2 - 10x + 25$ can be factored as $(x - 5)^2$.

Step 2: Find the critical points by setting the expression equal to zero and solving for $x$. Setting $(x - 5)^2 = 0$ and solving for $x$ gives us $x = 5$.

Step 3: Analyze the intervals: We have one critical point, $x = 5$. Now, we need to check the inequality in three intervals: $x < 5$, $x = 5$, and $x > 5$.

Step 4: Test the inequality in each interval:

For $x < 5$: Let's pick $x = 0$ (a value less than 5) and check the inequality: $(0 - 5)^2 = 25 > 0$

For $x = 5$: We already know that $(5 - 5)^2 = 0$, but the inequality is strict, so it does not include this point.

For $x > 5$: Let's pick $x = 6$ (a value greater than 5) and check the inequality: $(6 - 5)^2 = 1 > 0$

Step 5: Determine the solution: The inequality $x^2 - 10x + 25 < 0$ is satisfied for $x$ values less than 5 and greater than 5. In interval notation, the solution is:

$x \in (-\infty, 5) \cup (5, \infty)$