Question

1-x^2+8x+7 больше нуля 2-x^2-8x+16 больше нуля 3-2x^2-(5/2)x+1 больше нуля

Answer 1

Решение на фотографии.

Answer 2

To determine the values of x for which each expression is greater than zero, we need to find the solutions to the corresponding quadratic inequalities. Let's solve each one separately:

1. $1 - x^2 + 8x + 7 > 0$:

To solve this inequality, we can factor the quadratic expression: $1 - x^2 + 8x + 7 = -(x^2 - 8x - 8)$

Now, let's factor the quadratic expression inside the parentheses: $-x^2 + 8x - 8 = -(x - 4)(x + 2)$

To determine the sign of the expression, we can use a sign chart or test points. Let's use the latter approach. We will test three values: x = -3, x = 0, and x = 5.

For x = -3: $-(-3 - 4)(-3 + 2) = 7 \cdot -1 = -7 < 0$

For x = 0: $-(-4)(2) = 8 > 0$

For x = 5: $- (5 - 4)(5 + 2) = -1 \cdot 7 = -7 < 0$

The expression is greater than zero between the roots of the quadratic, which are x = -2 and x = 4. So the solution is: $x \in (-2, 4)$

2. $2 - x^2 - 8x + 16 > 0$:

Similarly, let's factor the quadratic expression: $2 - x^2 - 8x + 16 = -(x^2 + 8x - 14)$

Factoring the quadratic expression inside the parentheses: $-x^2 - 8x + 14 = -(x + 7)(x - 2)$

Let's test some values:

For x = -8: $-(-8 + 7)(-8 - 2) = 1 \cdot (-10) = -10 < 0$

For x = 0: $-7 \cdot 2 = -14 < 0$

For x = 3: $-(3 + 7)(3 - 2) = -10 < 0$

The expression is greater than zero between the roots of the quadratic, which are x = -7 and x = 2. So the solution is: $x \in (-7, 2)$

3. $3 - 2x^2 - \frac{5}{2}x + 1 > 0$:

Let's simplify the equation first: $3 - 2x^2 - \frac{5}{2}x + 1 = -2x^2 - \frac{5}{2}x + 4$

Now, we need to factor the quadratic expression: $-2x^2 - \frac{5}{2}x + 4$

Since the quadratic expression doesn't factor nicely, we can use the quadratic formula to find the roots: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For this equation, a = -2, b = -5/2, and c = 4.

Substituting these values into the quadratic formula:

$x = \frac{-\left(-\frac{5}{2}\right) \pm \sqrt{\left(-\frac{5}{2}\right)^2 - 4(-2)(4)}}{2(-2)}$

Simplifying:

$x = \frac{\frac{5}{2} \pm \sqrt{\frac{25}{4} + 32}}{-4}$ $x = \frac{\frac{5}{2} \pm \sqrt{\frac{25 + 128}{4}}}{-4}$ $x = \frac{\frac{5}{2} \pm \sqrt{\frac{153}{4}}}{-4}$