A. 3x + 5 = 0 B. 2x²+ 3x - 5 = 0 C. 4x³ - 2x²+ 1 = 0 D. 1/x + 2 = 0 相关知识点: 试题来源: 解析 B。解析:二次方程(quadratic equation)的一般形式是ax²+bx + c = 0(a≠0),选项B符合这个形式。选项A是一次方程。选项C是三次方程,因为最高次项是x³。选项D...
百度试题 结果1 题目Which quadratic equation has exactly one real root? ( ) A. -9x^2-6x+1=0 B. 9x^2-6x-1=0 C. 9x^2-6x+1=0 D. 9x^2+6x-1=0 相关知识点: 试题来源: 解析 C 反馈 收藏
Which of the following is the equation of a quadratic function with a vertex at (2,3) and opening upward? A. y = (x - 2)² + 3 B. y = (x + 2)² + 3 C. y = -(x - 2)² + 3 D. y = -(x + 2)² + 3 相关知识点: ...
百度试题 结果1 题目MULTIPLE CHOICE Which quadratic equation has the solutions x=(-9±√(81-56))/4 A 2x^2+9x-7=0③ 2x^2-9x+7=0⊙4x^2+9x+14=0 2x^2+9x+7=0 相关知识点: 试题来源: 解析 D 反馈 收藏
According to the problem, coefficients of the required quadratic equation arereal and its one root is -2 + i.We know in a quadratic with real coefficients imaginary roots occur inconjugate pairs).Since equation has rational coefficients, the other root is -2 - iNow, the sum of the roots ...
ait is 54.5 metres tall 它是54.5米高的[translate] a我立即意识到抓紧现在的一分一秒是多么至关重要 I realize the push present a minute second am immediately how very important[translate] awrinkle reducer 皱痕还原剂[translate] asolve this quadratic equation and hence find the pairs (x,y) for w...
结果1 题目 Using the Quadratic Formula, which of the following is the solution to the quadratic equation 3x^2+5x-12=0? ( ) A. x=-3, x=43 B. x=3, x=43 C. x=3, x=-43 D. x=-3, x=-43 相关知识点: 试题来源: 解析 A 反馈 收藏 ...
所以a=-3 a+1=-2 两个解是-2 -3 分析总结。 一题简单的英文版高级数学题结果一 题目 一题简单的英文版高级数学题.The quadratic equation x^2-x=k(x+1) has non-zero roots which differ by 1.Find the value of (i) each root,(ii)the constant k.翻译:二次方程 x^2-x=k(x+1)具有差别...
这兴趣比i的movieis它是。 [translate] a謝謝聆聽 Thanks listens respectfully [translate] aQuadratic equation (polynomial of second degree) was considered for the location of overturning moment arm of the total driving resultant (see Figure 5.34). 二次方程(第二级多项式)为翻转驾驶结果的共计的力矩臂...
Then, given equation becomes, y2−y+log16k=0 This is a quadratic equation and it has exactly one solution. ∴D=0 ⇒b2−4ac=0 ⇒(−1)2−4(1)(log16k)=0 ⇒log16k=14 ⇒k=1614 ⇒k=2 So, there is only one value for k. So, option (b) is the correct option. ...