2 2x 4



2 2x 4

Solve 2x^2=x4, Microsoft Math Solver

Add 4 to both sides. 2x2 x + 4 = 0 This equation is in standard form: ax2 + bx +c = 0. Substitute 2 for a, 1 for b, and 4 for c in the quadratic formula, 2ab± b24ac

Solve x^22x+4=0, Microsoft Math Solver

All equations of the form a x 2 + b x + c = 0 can be solved using the quadratic formula: 2 a b ± b 2 4 a c . The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction. x^ {2}2x+4=0. x 2 2 x + 4 = 0. This equation is in standard form: ax^ {2}+bx+c=0. Substitute 1 for a, 2 for b, and 4

Solve for x 2(2x4)=3(2x4), Mathway

Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with stepbystep explanations, just like a math tutor.

x^22x+4=0怎么解 百度知道

x^22x+11+4=0 x^22x+1+3=0 (x1)^2+3=0 (x1)^2=3 x1=正负根号3i 所以,x1=1+根号3i,x2=1根号3i 如果你学过复数,那么就是上面的答案,如果你还是初中生,那么,这道题目错了 已赞过 已踩过 你对这个回答的评价是? 评论 收起 chenghy666

Solve 4x^22x2=0, Microsoft Math Solver

4 { x }^ { 2 } 2x2 = 0. 4x2 2x 2 = 0. Divide both sides by 2. Divide both sides by 2. 2x^ {2}x1=0. 2x2 x 1 = 0. To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^ {2}+ax+bx1. To find a and b, set up a system to be solved.

2x4=20 solution

Simple and best practice solution for 2x4=20 equation. Check how easy it is, and learn it for the future. Our solution is simple, and easy to understand, so don`t hesitate to use it as a solution of your homework.

Solve x^3+2x^2+2x+4, Microsoft Math Solver

Do the grouping x^{3}+2x^{2}+2x+4=\left(x^{3}+2x^{2}\right)+\left(2x+4\right), and factor out x^{2} in the first and 2 in the second group.

分解因式:2x^4x^36x^2x+2

分解因式2x^4 3x^3+6x^2 6x+8 —— 解 先试设2x^43x^3+6x^26x+8=(x^2+ax+2)(2x^2+bx+4)=2x^4+(2a+b)x^3+(2*2+4+ab)x^2+(4a+2b)x+8 ∴2a+b=3①,ab+8=6>ab=2②,4a+2b=6>2a+b=3③ ①与③相同 ∴只要联立解出①,②即可 由2a+b=3>b=2a3④ ④

阅读材料:把形如ax2+bx+c的二次三项式(或其一部分)配成

是x 22x+4的三种不同形式的配方(即“余项”分别是常数项、一次项、二次项见横线上的部分). 请根据阅读材料解决下列问题: (1)比照上面的例子,写出x 2 +3x+1三种不同形式的配方; (2)将a 2 +ab+b 2 配方(至少两种形式); (3)已知a 2 +b 2 +c 2ab3b2c+4=0,求a+b+c的值.

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