Example Question - quadratic roots
Here are examples of questions we've helped users solve.
Solving Quadratic Equations Using the Quadratic Formula
题目要求解方程 \(x^2 - 5x + 5 = 0\)。我们可以用求根公式（也称为二次方程的根的通用公式）来解这个方程。该公式为： \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] 其中，\(a\)、\(b\) 和 \(c\) 分别是方程 \(ax^2 + bx + c = 0\) 中的系数。在我们的例子中，\(a = 1\)、\(b = -5\)、\(c = 5\)。 根据这些值，我们可以将它们代入公式： \[x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1}\] \[x = \frac{5 \pm \sqrt{25 - 20}}{2}\] \[x = \frac{5 \pm \sqrt{5}}{2}\] 所以，方程的解是： \[x_1 = \frac{5 + \sqrt{5}}{2}\] \[x_2 = \frac{5 - \sqrt{5}}{2}\] 这就是方程 \(x^2 - 5x + 5 = 0\) 的两个解。
Solving Quadratic Equations with the General Formula
Para resolver la ecuación cuadrática \( x^2 + 8x - 5 \), primero intentaremos encontrar dos números que multiplicados den como resultado -5 y que sumados den como resultado 8. Sin embargo, en este caso no es fácil encontrar dos números que cumplan con estas condiciones, por lo que usaré la fórmula general para resolver ecuaciones cuadráticas, que es: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] En la ecuación dada, \( a = 1 \), \( b = 8 \), y \( c = -5 \). Sustituyendo estos valores en la fórmula general, obtenemos: \[ x = \frac{-8 \pm \sqrt{8^2 - 4(1)(-5)}}{2(1)} \] \[ x = \frac{-8 \pm \sqrt{64 + 20}}{2} \] \[ x = \frac{-8 \pm \sqrt{84}}{2} \] \[ x = \frac{-8 \pm 2\sqrt{21}}{2} \] \[ x = -4 \pm \sqrt{21} \] Por lo tanto, las soluciones de la ecuación cuadrática son \( x = -4 + \sqrt{21} \) y \( x = -4 - \sqrt{21} \). Estas son las dos raíces de la función cuadrática y los puntos en los cuales la curva de la función cortará el eje x en una gráfica.
Solving Quadratic Equations with Quadratic Formula
The image contains a mathematical expression: \[ 3x^2 + x - 10 \] To solve this quadratic equation, we would typically first try to factor it, if possible. However, if the quadratic doesn't factor easily or at all, then we can use the quadratic formula to find the roots of the equation, which is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] In this case, the coefficients for our quadratic expression \(3x^2 + x - 10\) are \(a = 3\), \(b = 1\), and \(c = -10\). Let's apply the quadratic formula: \[ x = \frac{-1 \pm \sqrt{1^2 - 4(3)(-10)}}{2(3)} \] \[ x = \frac{-1 \pm \sqrt{1 + 120}}{6} \] \[ x = \frac{-1 \pm \sqrt{121}}{6} \] \[ x = \frac{-1 \pm 11}{6} \] This gives us two solutions: \[ x = \frac{-1 + 11}{6} = \frac{10}{6} = \frac{5}{3} \] \[ x = \frac{-1 - 11}{6} = -2 \] Therefore, the two solutions to the equation \(3x^2 + x - 10 = 0\) are \(x = \frac{5}{3}\) and \(x = -2\).
Understanding Quadratic Equations
An equation that can be written in the form \( ax^2 + bx + c = 0 \) (with \( a \neq 0 \)) is called a quadratic equation.
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