## Applying the Quadratic Formula The quadratic formula is a powerful tool for solving quadratic equations, which are equations of the form $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are constants and $a \neq 0$. While factoring can solve some quadratic equations, the quadratic formula works for *all* quadratic equations, even those that are not easily factorable. Remember the quadratic formula is your friend when factoring fails! ### The Quadratic Formula The quadratic formula itself is: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ This formula gives you the two possible solutions (roots) for $x$ in the quadratic equation. The $\pm$ symbol indicates that there are two solutions: one where you add the square root term and one where you subtract it. The expression inside the square root, $b^2 - 4ac$, is called the discriminant. It tells you about the nature of the solutions (real or complex, and how many). ### Steps to Apply the Quadratic Formula Solving a quadratic equation using the quadratic formula involves three key steps: **Step 1: Identify a, b, and c** First, identify the values of $a$, $b$, and $c$ in your quadratic equation. Make sure your equation is in the standard form ($ax^2 + bx + c = 0$) before you begin. In the equation $2x^2 - 5x + 3 = 0$, we have: * $a = 2$ * $b = -5$ * $c = 3$ **Step 2: Substitute into the Formula** Next, substitute the values of $a$, $b$, and $c$ into the quadratic formula. Be careful with signs – negative numbers can be tricky! Using the values from the previous example, we substitute into the formula: $$x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(3)}}{2(2)}$$ **Step 3: Simplify** Finally, simplify the expression using the order of operations (PEMDAS/BODMAS). This often involves simplifying the expression under the square root first, then performing the addition and subtraction, and finally dividing by $2a$. Continuing from the previous example: $$x = \frac{5 \pm \sqrt{25 - 24}}{4}$$ $$x = \frac{5 \pm \sqrt{1}}{4}$$ $$x = \frac{5 \pm 1}{4}$$ This gives us two solutions: $x = \frac{5 + 1}{4} = \frac{6}{4} = \frac{3}{2}$ $x = \frac{5 - 1}{4} = \frac{4}{4} = 1$ Therefore, the solutions to the equation $2x^2 - 5x + 3 = 0$ are $x = \frac{3}{2}$ and $x = 1$. ### Example Problem 1: Solve $5x^2 - 9x + 2 = 0$ using the quadratic formula. 1. **Identify a, b, c:** $a = 5$, $b = -9$, $c = 2$ 2. **Substitute:** $x = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(5)(2)}}{2(5)}$ 3. **Simplify:** $x = \frac{9 \pm \sqrt{81 - 40}}{10} = \frac{9 \pm \sqrt{41}}{10}$ The solutions are $x = \frac{9 + \sqrt{41}}{10}$ and $x = \frac{9 - \sqrt{41}}{10}$. ### Example Problem 2: Solve $3x^2 + 8x - 7 = 0$ using the quadratic formula. 1. **Identify a, b, c:** $a = 3$, $b = 8$, $c = -7$ 2. **Substitute:** $x = \frac{-8 \pm \sqrt{8^2 - 4(3)(-7)}}{2(3)}$ 3. **Simplify:** $x = \frac{-8 \pm \sqrt{64 + 84}}{6} = \frac{-8 \pm \sqrt{148}}{6} = \frac{-8 \pm 2\sqrt{37}}{6} = \frac{-4 \pm \sqrt{37}}{3}$ The solutions are $x = \frac{-4 + \sqrt{37}}{3}$ and $x = \frac{-4 - \sqrt{37}}{3}$. A common mistake is forgetting to change the sign of 'b' when substituting into the formula. Always double-check your signs! Practice is key to mastering the quadratic formula. Try solving different quadratic equations to build your confidence and understanding.