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All equations of the form ax^ {2}+bx+c=0 can be solved using the quadratic formula: \frac {-b±\sqrt {b^ {2}-4ac}} {2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction. Now solve the equation x=\frac {4±2\sqrt {3}} {2} when ± …

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2014年1月9日 · x²－4x＋1＝0的配方法解答过程如下：（1）把1化成4-3：x²-4x+4-3=0（2）x²-4x+4可以组成一个完全平方式： (x-2)²=3（3）两边同时开根号.

形式为 ax^ {2}+bx+c=0 的所有方程式均可求解，方法是使用二次公式来求解: \frac {-b±\sqrt {b^ {2}-4ac}} {2a}。 此二次公式可得到两个解，一个是当 ± 取加号时的解，另一个是取减号时的解。 现在 ± 为加号时求公式 x=\frac {4±2\sqrt {3}} {2} 的解。 将 2\sqrt {3} 加上 4。 现在 ± 为减号时求公式 x=\frac {4±2\sqrt {3}} {2} 的解。 将 4 减去 2\sqrt {3}。 使用 ax^ {2}+bx+c=a\left (x-x_ {1}\right)\left (x-x_ {2}\right) 对原始表达式进行因式分解。

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All equations of the form ax^ {2}+bx+c=0 can be solved using the quadratic formula: \frac {-b±\sqrt {b^ {2}-4ac}} {2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction. This equation is in standard form: ax^ {2}+bx+c=0.

x^2: x^{\msquare} \log_{\msquare} \sqrt{\square} \nthroot[\msquare]{\square} \le \ge \frac{\msquare}{\msquare} \cdot \div: x^{\circ} \pi \left(\square\right)^{'} \frac{d}{dx} \frac{\partial}{\partial x} \int \int_{\msquare}^{\msquare} \lim \sum \infty \theta (f\:\circ\:g) f(x)

解：∵x2﹣4x﹣1=0， ∴x2﹣4x=1，∴x2﹣4x+4=1+4，∴（x﹣2）2=5，∴x=2± ，∴x1=2+ ，x2=2﹣ 【分析】配方法的一般步骤：（1）把常数