Full solution

Q. Solve using the quadratic formula.

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8f^2 + 4f - 1 = 0

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Write your answers as integers, proper or improper fractions in simplest form, or decimals rounded to the nearest hundredth.

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f =

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f =

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Identify coefficients: Identify the coefficients $a$ , $b$ , and $c$ in the quadratic equation $8f^2 + 4f - 1 = 0$ by comparing it to the standard form $ax^2 + bx + c = 0$ . $\newline$ $a = 8$ $\newline$ $b = 4$ $\newline$ $c = -1$
Substitute values into formula: Substitute the values of $a$ , $b$ , and $c$ into the quadratic formula $f = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ . $f = \frac{-(4) \pm \sqrt{(4)^2 - 4\cdot(8)\cdot(-1)}}{2\cdot(8)}$
Simplify discriminant: Simplify the expression under the square root (the discriminant). $\sqrt{(4)^2 - 4\cdot(8)\cdot(-1)} = \sqrt{16 + 32} = \sqrt{48}$
Simplify square root: Further simplify the square root of $48$ to its simplest radical form. $\newline$ $\sqrt{48} = \sqrt{16 \times 3} = 4 \times \sqrt{3}$
Substitute simplified value: Now, substitute the simplified square root back into the quadratic formula. $\newline$ $f = (-(4) \pm 4 \cdot \sqrt{3}) / (2\cdot(8))$ $\newline$ $f = (-4 \pm 4 \cdot \sqrt{3}) / 16$
Divide by denominator: Simplify the expression by dividing both terms in the numerator by the denominator. $\newline$ $f = \left(-\frac{1}{4} \pm \left(\frac{1}{4}\right) \cdot \sqrt{3}\right)$
Find two solutions: Now, we have two possible solutions for $f$ . $f = -\frac{1}{4} + \left(\frac{1}{4}\right) \cdot \sqrt{3}$ or $f = -\frac{1}{4} - \left(\frac{1}{4}\right) \cdot \sqrt{3}$
Round to nearest hundredth: If necessary, round the values of $f$ to the nearest hundredth. $f \approx -0.25 + 0.25 \times 1.73$ or $f \approx -0.25 - 0.25 \times 1.73$ $f \approx -0.25 + 0.4325$ or $f \approx -0.25 - 0.4325$ $f \approx 0.1825$ or $f \approx -0.6825$ $f \approx 0.18$ or $f \approx -0.68$