We are given the quadratic inequality:

$8x^2 + 2x - 3 \geq 0$

Step 1: Apply the Quadratic Formula

The quadratic formula is:

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

For our equation, the coefficients are:

• $a = 8$
• $b = 2$
• $c = -3$

Step 2: Compute the Discriminant

The discriminant is given by:

$b^2 - 4ac = (2)^2 - 4(8)(-3)$

$= 4 + 96 = 100$

Since the discriminant is a perfect square, we can proceed with solving for $x$.

Step 3: Compute the Roots

$x = \frac{-2 \pm \sqrt{100}}{2(8)}$

$= \frac{-2 \pm 10}{16}$

Solving for each root:

$x_1 = \frac{-2 + 10}{16} = \frac{8}{16} = \frac{1}{2}$

$x_2 = \frac{-2 - 10}{16} = \frac{-12}{16} = \frac{-3}{4}$

Step 4: Write the Equivalent Factored Form

Using these roots, the factors of the quadratic expression can be written as:

$(x - x_1)(x - x_2) = (x - \frac{1}{2})(x + \frac{3}{4})$

Since the original quadratic expression has a leading coefficient of 8, we multiply the factors by 8:

$8(x - \frac{1}{2})(x + \frac{3}{4}) \geq 0$

Rewriting in a cleaner form:

$(8x - 4)(x + \frac{3}{4}) \geq 0$

Step 5: Find Intervals to Test

The roots divide the number line into three intervals:

1. $(-\infty, -\frac{3}{4})$
2. $(-\frac{3}{4}, \frac{1}{2})$
3. $(\frac{1}{2}, \infty)$

To determine where the expression is nonnegative, we test one point from each interval.

1. Test $x = -1$ in $(-\infty, -\frac{3}{4})$:
   $(8(-1) - 4)((-1) + \frac{3}{4}) = (-8 - 4)(-1 + 0.75) = (-12)(-0.25) = 3$ Positive ✅

2. Test $x = 0$ in $(-\frac{3}{4}, \frac{1}{2})$:
   $(8(0) - 4)((0) + \frac{3}{4}) = (-4)(0.75) = -3$ Negative ❌

3. Test $x = 1$ in $(\frac{1}{2}, \infty)$:
   $(8(1) - 4)((1) + \frac{3}{4}) = (8 - 4)(1.75) = (4)(1.75) = 7$ Positive ✅

Step 6: Write the Final Solution

Since we need the values where $8x^2 + 2x - 3 \geq 0$, we include the intervals where the expression is positive:

$x \in (-\infty, -\frac{3}{4}] \cup [\frac{1}{2}, \infty)$

Since the inequality is greater than or equal to 0, we include the roots $x = -\frac{3}{4}$ and $x = \frac{1}{2}$, where the expression equals 0.

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Would you like a more detailed explanation for any step?

Related Questions:

1. How does the quadratic formula help solve quadratic inequalities?
2. What is the significance of the discriminant in determining the nature of solutions?
3. How can you determine the signs of factors in different intervals?
4. Why do we multiply the factors by the leading coefficient when rewriting the expression?
5. How does testing intervals help in solving inequalities?

Tip: When solving quadratic inequalities, always check the boundary points separately to see if they satisfy the inequality.