Question

X^2-3x+4>0 и x^2-16<0​

Answer 1

${x}^{2} - 3x + 4 > 0 \\ x = r$

${x}^{2} - 16 < 0 \\ ( x + 4)(x - 4) < 0 \\ x > - 4 \\ x < 4$

Answer 2

To solve the inequalities $x^2 - 3x + 4 > 0$ and $x^2 - 16 < 0$, let's break them down one by one:

1. $x^2 - 3x + 4 > 0$:

To solve this quadratic inequality, we can factor the quadratic expression or use the quadratic formula to find the roots. However, since the inequality asks for where the expression is greater than 0, we can find the critical points and test intervals.

Let's find the roots of the equation $x^2 - 3x + 4 = 0$:

The roots are given by the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where the coefficients are $a = 1$, $b = -3$, and $c = 4$.

$x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}$

$x = \frac{3 \pm \sqrt{9 - 16}}{2}$

$x = \frac{3 \pm \sqrt{-7}}{2}$

Since the discriminant ($-7$) is negative, the quadratic has no real roots. This means the quadratic expression is either entirely positive or entirely negative.

To determine the intervals where $x^2 - 3x + 4 > 0$, we can use the fact that the coefficient of $x^2$ is positive, indicating that the parabola opens upwards.

However, since there are no real roots, the expression does not change sign. So, it is either always positive or always negative. In this case, $x^2 - 3x + 4$ is always positive ($> 0$).

2. $x^2 - 16 < 0$:

Let's solve this quadratic inequality in a similar manner:

$x^2 - 16 = 0$ can be factored as $(x - 4)(x + 4) = 0$, which gives us two real roots: $x = 4$ and $x = -4$.

Now, let's test the intervals created by these roots:

• When $x < -4$, both factors $(x - 4)$ and $(x + 4)$ are negative, so the whole expression is positive. Not in the solution.
• When $-4 < x < 4$, $(x - 4)$ is negative, but $(x + 4)$ is positive, making the expression negative. This interval is in the solution.
• When $x > 4$, both factors are positive, so the expression is positive. Not in the solution.

Therefore, the solution to $x^2 - 16 < 0$ is $-4 < x < 4$.

To summarize:

• For $x^2 - 3x + 4 > 0$, the solution is all real values of $x$ (the expression is always positive).
• For $x^2 - 16 < 0$, the solution is $-4 < x < 4$.