Solving quadratic equations can be efficiently done using the quadratic formula, which provides solutions for any quadratic equation in standard form \(ax^2 + bx + c = 0\). The quadratic formula is expressed as:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
This formula calculates the roots of the quadratic equation by substituting the coefficients \(a\), \(b\), and \(c\) directly into the formula. The term under the square root, \(b^2 - 4ac\), is called the discriminant and determines the nature of the solutions. If the discriminant is positive, there are two distinct real solutions; if it is zero, there is exactly one real solution; and if it is negative, the solutions are complex or imaginary.
For example, consider the quadratic equation \$2x^2 - 3x - 5 = 0\(. Here, \)a = 2\(, \)b = -3\(, and \)c = -5\(. Plugging these values into the quadratic formula gives:
\[x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-5)}}{2(2)} = \frac{3 \pm \sqrt{9 + 40}}{4} = \frac{3 \pm \sqrt{49}}{4} = \frac{3 \pm 7}{4}\]
This results in two solutions:
\[x = \frac{3 + 7}{4} = \frac{10}{4} = \frac{5}{2} \quad \text{and} \quad x = \frac{3 - 7}{4} = \frac{-4}{4} = -1\]
In another example, the quadratic \)x^2 - 8x + 16 = 0\( has \)a = 1\(, \)b = -8\(, and \)c = 16\(. Substituting these values yields:
\[x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(16)}}{2(1)} = \frac{8 \pm \sqrt{64 - 64}}{2} = \frac{8 \pm 0}{2} = \frac{8}{2} = 4\]
Here, the discriminant is zero, so there is only one real solution, \)x = 4\(.
The quadratic formula is a powerful tool because it works regardless of whether the coefficient \)a\( is one or any other number, and it handles cases where the coefficient \)b\( is odd or even. The plus-or-minus symbol (\)\pm$) in the formula accounts for the two possible solutions when the discriminant is positive. Always remember to verify your solutions by substituting them back into the original quadratic equation to ensure accuracy.
