roots of x2+px+m=0 are a and b, then using Vieta's equations, 2a+2b=−ma+b=−p2a(2b)=na(b)=m Therefore, substituting the second equation into the first equation gives m=2(p) and substituting the fourth equation into the third equation gives n=4(m)Therefore, n=8p, so np...
Indeed, consider the quadratics x2+8x+16=0, x2+16x+64=0. If the roots of x2+mx+n=0 are 2a and 2b and the roots of x2+px+m=0 are a and b, then using Vieta's equations, 2a+2b=−ma+b=−p2a(2b)=na(b)=m Therefore, substituting the second equation into the first ...
解：设y=a(x-1)²+2，把(3,10)代入上式，解得y=2(x-1)²+2。注意：与点在平面直角坐标系中的平移不同，二次函数平移后的顶点式中，h>0时，h越大，图像的对称轴离y轴越远，且在x轴正方向上，不能因h前是负号就简单地认为是向左平移。具体可分为下面几种情况：当h>0时，y=a(x-h)²...
NAGEEN PRAKASHAN-COMPLEX NUMBERS AND QUADRATIC EQUATION -EXERCISE 5.3 Solve the equation:x^2+3=0 01:43 Solve the equation:2x^2+x+1=0 01:41 Solve the equation:x^2+3x+9=0 01:55 Solve the equation:x^2-x+2=0 01:56 Solve the equation:x^2+3x+5=0 01:25 Solve the equation:x^2...
x^2-x=k(x+1)即x^2-(k+1)x-k=0可设两个为a,a+1那么a(a+1)=-ka+a+1=k+1两个式子联立消去k,得到a^3+3a=0由于是非零解所以a=-3a+1=-2两个解是-2 -3相关推荐 1一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 ...
Quadratic equation x2−4x−5=0 Trigonometry 4sinθcosθ=2sinθ Linear equation y=3x+4 Arithmetic 699∗533 Matrix [2534][2−10135] Simultaneous equation {8x+2y=467x+3y=47 Differentiation dxd(x−5)(3x2−2) Integration ∫01xe−x2dx Limits x→−3limx2+2x−3x2−9 Back...
x2−63x k=0 are prime numbers. The number of possible values of k is( ).A. 0 B. 1 C. 2 D. 4 E. more than 4 相关知识点: 试题来源: 解析 B Consider a general quadratic with the coefficient of x1 being 1 and the roots being r and s. It can be factored as (x−r)(...
The quadratic equation x^(2) + (a^(2) - 2) x - 2a^(2) and x^(2) - 3x + 2 = 0 have
The solvability of these diophantine equations are related (by a theorem of Moser) to the stufe (the minimal number of squares 1 i the sum of integral squares) of an imaginary quadratic number field. We obtain an explicit result that the stufe is 2. Finally, we easily prove some results ...