Question

solve by the quadratic formula. list the solutions.

\\-8x^2 + 6x - 1 = 0\\

\\(x = \\)

Explanation:

⚡ Using what you learned: quadratic formula and its applications

Step 1: Identify coefficients

Identify the coefficients \(a\), \(b\), and \(c\) from the standard form equation \(ax^2 + bx + c = 0\):

$$ -8x^2 + 6x - 1 = 0 $$

$$ a = -8, \quad b = 6, \quad c = -1 $$

Step 2: Apply the quadratic formula

Substitute the values into the quadratic formula:

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

$$ x = \frac{-(6) \pm \sqrt{(6)^2 - 4(-8)(-1)}}{2(-8)} $$

Step 3: Simplify the expression

Simplify the terms inside and outside the square root:

$$ x = \frac{-6 \pm \sqrt{36 - 32}}{-16} $$

$$ x = \frac{-6 \pm \sqrt{4}}{-16} $$

$$ x = \frac{-6 \pm 2}{-16} $$

Step 4: Calculate the two solutions

Evaluate the two possible cases:

Case 1 (plus):

$$ x = \frac{-6 + 2}{-16} = \frac{-4}{-16} = \frac{1}{4} $$

Case 2 (minus):

$$ x = \frac{-6 - 2}{-16} = \frac{-8}{-16} = \frac{1}{2} $$

Answer:

1/4, 1/2