Problem
\( 4: 59 \) く飞书 Result mathpix Api 请求Al参数 Sure! This is a quadratic equation in the form of \( a x^{2} \) \( +b x+c=0 \). In order to solve it, we need to use the quadratic formula: \( x=\left(-b \pm \operatorname{sqrt}\left(b^{2}-4 a c\right)\right) / 2 a \) Here are the steps to solve the equation: 1. Identify the values of \( a, b \), and \( c \). These are the coefficients of the \( \mathrm{x}^{2} \) term, the \( x \) term, and the constant term, respectively. 2. Plug these values into the quadratic formula and simplify: \[ \begin{array}{l} x=\left(-b \pm \operatorname{sqrt}\left(b^{2}-4 a c\right)\right) / 2 a \\ x=\left(-b \pm \operatorname{sqrt}\left(b^{2}-4 a c\right)\right) /(2 a) \end{array} \] 3. Simplify the expression under the square root, if possible. This is called the discriminant: \( b^{2}-4 a c \) 4. If the discriminant is positive, there are two real roots. If it's zero, there is one real root. If it's negative, there are two complex roots (which are more complicated to work with). 5. Plug the values of \( a, b \), and \( c \) into the quadratic formula and simplify: \[ x=\left(-b \pm \operatorname{sqrt}\left(b^{2}-4 a c\right)\right) / 2 a \] 6. If there are two real roots, write them as \( x= \) (root1, root2). If there is one real root, write it as \( x= \) (root). If there are two complex roots, write them as \( x= \) (real part \pm imaginary part). That's it! You've solved the quadratic equation.
Solution
Step 1 ：Identify the values of a, b, and c. These are the coefficients of the x^2 term, the x term, and the constant term, respectively.
Step 2 ：Plug these values into the quadratic formula and simplify: $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$
Step 3 ：Substitute a=1, b=-2, and c=4 into the formula.
Step 4 ：Simplify: $x=\frac{-(-2) \pm \sqrt{(-2)^2-4(1)(4)}}{2(1)}$
Step 5 ：Simplify further: $x=\frac{2 \pm \sqrt{4(4)}}{2}$
Step 6 ：More simplification: $x=\frac{-1 \pm \sqrt{3}}{3}$
Step 7 ：Final answer: $x=\frac{-1 + \sqrt{3}}{3}$ or $x=\frac{-1 - \sqrt{3}}{3}$
