Question

solve for x using the quadratic formula.

\\2x^2 - 18x + 16 = 0\\

\\x = ?, \quad \\

enter the smallest solution first.

quadratic formula: \\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\)

Explanation:

Identify coefficients

We identify the coefficients from the given quadratic equation.
Using the Solving Quadratic Functions knowledge point

$$ 2x^2 - 18x + 16 = 0 \implies a = 2, \quad b = -18, \quad c = 16 $$

Apply quadratic formula

We substitute the coefficients into the quadratic formula.

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

$$ x = \frac{-(-18) \pm \sqrt{(-18)^2 - 4(2)(16)}}{2(2)} $$

Simplify the radicand

We calculate the value inside the square root.

$$ b^2 - 4ac = 324 - 128 = 196 $$

$$ \sqrt{196} = 14 $$

Solve for x

We compute both possible values of the variable.

$$ x = \frac{18 \pm 14}{4} $$

$$ x_1 = \frac{18 - 14}{4} = 1, \quad x_2 = \frac{18 + 14}{4} = 8 $$

Order the solutions

We arrange solutions from smallest to largest.

$$ x = 1, 8 $$

Answer:

Solve for x using the quadratic formula.

$$2x^2 - 18x + 16 = 0$$

\(x =\) <blank>1</blank>, <blank>8</blank>