Standard Representation: Using roots

To solve the problem, let's determine when $y = x^2 + \frac{1}{2}x - 4.5$ is greater than zero by following these steps:

• Step 1: Find the roots of the quadratic equation using the quadratic formula.
• Step 2: Determine the intervals defined by these roots.
• Step 3: Identify where the quadratic function is positive.

**Step 1**: Given the quadratic function $y = x^2 + \frac{1}{2}x - 4.5$, the coefficients are $a = 1$, $b = \frac{1}{2}$, and $c = -4.5$. Apply the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-\left(\frac{1}{2}\right) \pm \sqrt{\left(\frac{1}{2}\right)^2 - 4(1)(-4.5)}}{2(1)}$

Simplify further:

$x = \frac{-\frac{1}{2} \pm \sqrt{\frac{1}{4} + 18}}{2} = \frac{-\frac{1}{2} \pm \sqrt{\frac{73}{4}}}{2}$

This becomes:

$x = \frac{-1 \pm \sqrt{73}}{4}$

**Step 2**: The roots are $x_1 = \frac{-1 - \sqrt{73}}{4}$ and $x_2 = \frac{-1 + \sqrt{73}}{4}$. The function changes sign at the roots. Since the quadratic opens upwards (as $a = 1 > 0$), it will be positive between the roots:

**Step 3**: Identify the interval where the quadratic is positive:

$\frac{-1 - \sqrt{73}}{4} < x < \frac{-1 + \sqrt{73}}{4}$

Therefore, the values of $x$ for which the function is greater than zero are within this interval.

The correct solution is $\frac{-1 - \sqrt{73}}{4} < x < \frac{-1 + \sqrt{73}}{4}$.

To solve the problem, we need to determine the values of $x$ for which the quadratic function $y = -x^2 + 2\frac{1}{2}x - \frac{1}{4}$ is less than zero.

Let's start by solving the equation $-x^2 + \frac{5}{2}x - \frac{1}{4} = 0$ to find the roots using the quadratic formula:

The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = -1$, $b = \frac{5}{2}$, and $c = -\frac{1}{4}$.

Calculate the discriminant:

$b^2 - 4ac = \left(\frac{5}{2}\right)^2 - 4(-1)(-\frac{1}{4}) = \frac{25}{4} - 1 = \frac{21}{4}$.

Since the discriminant is positive, there are two distinct real roots.

Next, plug the discriminant back into the quadratic formula to find the roots:

$x = \frac{-\frac{5}{2} \pm \sqrt{\frac{21}{4}}}{2(-1)} = \frac{-\frac{5}{2} \pm \frac{\sqrt{21}}{2}}{-2} = \frac{5 \pm \sqrt{21}}{4}$.

Thus, the roots are $x = \frac{5 + \sqrt{21}}{4}$ and $x = \frac{5 - \sqrt{21}}{4}$.

The parabola opens downwards (since $a = -1$), so the function is positive between the roots and negative outside. Therefore, $f(x) < 0$ for $x > \frac{5+\sqrt{21}}{4}$ or $x < \frac{5-\sqrt{21}}{4}$.

The correct answer is $x > \frac{5+\sqrt{21}}{4}$ or $x < \frac{5-\sqrt{21}}{4}$.

To solve this problem, follow these steps:

• Step 1: Use the quadratic formula to find the roots of $y = -x^2 + \frac{5}{2}x - \frac{1}{4}$.
• Step 2: Determine the intervals based on these roots and check where the function is positive.

Step 1: Find the roots using the quadratic formula:
The quadratic equation is $-x^2 + \frac{5}{2}x - \frac{1}{4} = 0$, with $a = -1$, $b = \frac{5}{2}$, and $c = -\frac{1}{4}$.
The roots are given by:

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

Substitute the values into the formula:

$x = \frac{-\frac{5}{2} \pm \sqrt{\left(\frac{5}{2}\right)^2 - 4(-1)\left(-\frac{1}{4}\right)}}{2(-1)}$

$x = \frac{-\frac{5}{2} \pm \sqrt{\frac{25}{4} - 1}}{-2}$

$x = \frac{-\frac{5}{2} \pm \sqrt{\frac{21}{4}}}{-2}$

$x = \frac{-\frac{5}{2} \pm \frac{\sqrt{21}}{2}}{-2}$

Simplify each root:

$x = \frac{5 \pm \sqrt{21}}{4}$

Step 2: Determine the intervals:
The roots are $x = \frac{5-\sqrt{21}}{4}$ and $x = \frac{5+\sqrt{21}}{4}$.

• The function is a downward-opening parabola (because $a = -1$), so it is positive between the roots and negative outside them.

Therefore, $f(x) > 0$ for $\frac{5-\sqrt{21}}{4} < x < \frac{5+\sqrt{21}}{4}$.

The solution is $\frac{5-\sqrt{21}}{4} < x < \frac{5+\sqrt{21}}{4}$.

To solve this problem, we will analyze the quadratic function $y = x^2 + \frac{1}{2}x - 4\frac{1}{2}$. We need to determine for which values of $x$ the function $f(x) > 0$.

• Step 1: Identify the coefficients: The function can be rewritten in standard form as $y = ax^2 + bx + c$, where $a = 1$, $b = \frac{1}{2}$, $c = -\frac{9}{2}$.
• Step 2: Use the quadratic formula to find the roots: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Step 3: Calculate the discriminant: $\Delta = b^2 - 4ac = \left(\frac{1}{2}\right)^2 - 4(1)\left(-\frac{9}{2}\right) = \frac{1}{4} + 18 = \frac{73}{4}$. Since $\Delta > 0$, there are two real roots.
• Step 4: Find the roots: $x = \frac{-\frac{1}{2} \pm \sqrt{\frac{73}{4}}}{2} = \frac{-\frac{1}{2} \pm \frac{\sqrt{73}}{2}}{2}$ $x = \frac{-1 \pm \sqrt{73}}{4}$ These roots are $x_1 = \frac{-1 - \sqrt{73}}{4}$ and $x_2 = \frac{-1 + \sqrt{73}}{4}$.
• Step 5: Analyze the sign of the quadratic: The parabola opens upwards (as $a = 1 > 0$). It is positive outside its roots: $x < x_1$ and $x > x_2$.

Therefore, the values of $x$ that satisfy $f(x) > 0$ are $x > \frac{-1+\sqrt{73}}{4}$ or $x < \frac{-1-\sqrt{73}}{4}$.

The solution is $x > \frac{-1+\sqrt{73}}{4}$ or $x < \frac{-1-\sqrt{73}}{4}$.

To solve this problem, we will apply the following steps:

• Step 1: Find the roots of the given quadratic function.
• Step 2: Determine the sign of the quadratic between and beyond the roots.
• Step 3: Conclude which values of $x$ make the quadratic positive.

Let's start with Step 1: Find the roots of the function $y = x^2 - 4x - 4$.
We use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting $a = 1, b = -4, c = -4$, we get:
$\Delta = b^2 - 4ac = (-4)^2 - 4(1)(-4) = 16 + 16 = 32$

Thus, the roots are:
$x = \frac{-(-4) \pm \sqrt{32}}{2(1)} = \frac{4 \pm \sqrt{32}}{2}$

Simplifying further gives us $\sqrt{32} = 4\sqrt{2}$, so:
$x = \frac{4 \pm 4\sqrt{2}}{2} = 2 \pm 2\sqrt{2}$

The roots are $x = 2 + 2\sqrt{2}$ and $x = 2 - 2\sqrt{2}$.

Step 2: Determine the sign of the quadratic.
Since the parabola opens upward (coefficient of $x^2$ is positive), it is below the x-axis between the roots and above the x-axis outside the roots.

Step 3: Conclude values for which $f(x) > 0$.
$f(x) > 0$ for $x < 2 - 2\sqrt{2}$ or $x > 2 + 2\sqrt{2}$.

Finally, the solution to the problem is: $x > 2 + 2\sqrt{2}$ or $x < 2 - 2\sqrt{2}$.

We are tasked with finding the values of $x$ for which the function $y = x^2 - 4x - 4$ satisfies $f(x) < 0$.

To solve this, follow these steps:

• Find the roots of the function:
  The roots of a quadratic equation $ax^2 + bx + c = 0$ can be found using the quadratic formula: $\ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ For our function, $a = 1$, $b = -4$, and $c = -4$. Plug these into the quadratic formula: $\ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \times 1 \times (-4)}}{2 \times 1}$ $\ x = \frac{4 \pm \sqrt{16 + 16}}{2}$ $\ x = \frac{4 \pm \sqrt{32}}{2}$ $\ x = \frac{4 \pm 4\sqrt{2}}{2}$ $\ x = 2 \pm 2\sqrt{2}$ Thus, the roots are $x = 2 + 2\sqrt{2}$ and $x = 2 - 2\sqrt{2}$.
• Determine intervals:
  The function changes sign at these roots. The intervals to consider are $(-\infty, 2 - 2\sqrt{2})$, $(2 - 2\sqrt{2}, 2 + 2\sqrt{2})$, and $(2 + 2\sqrt{2}, \infty)$.
• Test each interval for $f(x) < 0$:
  Since the parabola opens upwards (as $a > 0$), the function will be negative between the roots: - In $(2 - 2\sqrt{2}, 2 + 2\sqrt{2})$, the function is less than zero. Thus, $f(x) < 0$ for $2 - 2\sqrt{2} < x < 2 + 2\sqrt{2}$.

Therefore, the solution is $\boxed{2 - 2\sqrt{2} < x < 2 + 2\sqrt{2}}$.