Hence the equation in the question becomes y^5-(32)(y^5)=31, i.e. (y^5-32)(y^5+1)=0. It follows that y may be 2 or -1. ●If x+2+√(x^2+4x+3) =2, we have x^2+4x+3 = (-x)^2, which gives x=-34. ●If x+2+√(x^2+4x+3) =-1, we have x^2+4x+...
Find all real roots of the equation (x= + 4)(2x − 1) = (x= + 4)(2x + 1) by a method that involves factoring. [ Show all work.] 相关知识点: 试题来源: 解析 ∅, There are no real rootsWORK SHOWN: (x= + 4)(2x − 1) = (x= + 4)(2x + 1), (x= + 4)(2x...
Find the real roots of the equation x3-9x+1=0 by method of false position.Show transcribed image text There are 4 steps to solve this one. Solution Share Step 1 Given equation is x3−9x+1=0 , we have to solve this equation by using false position method Expla...
(ii)Find all the roots of the equation z^(10)+z^5+1=0 ,giving each root in the form eie.[5] 相关知识点: 试题来源: 解析 3(ii) z^5=-1/2+i(√3)/2 z^5=erp(i2π(1/3+k)(i2π(-1/3+k)) exp(±t(2π)/(15),exp(t+(4π)/(15))⋅exp(t+(2π)/3),exp...
We have to find the roots of the given equation.Solution: 1x+4−1x−7=1130,x≠−4,71x+4−1x−7=1130,x≠−4,7 1x+4−1x−7=11301x+4−1x−7=1130 x−7−(x+4)(x+4)(x−7)=1130x−7−(x+4)(x+4)(x−7)=1130...
结果1 题目 f(z)=z^2+5z+10Find the roots of the equation f(z)=0, giving your answers in the form a± ib, where a and b are real numbers. 相关知识点: 试题来源: 解析 z=- 52+ (√ (15))2i and z=- 52- (√ (15))2i 反馈 收藏 ...
Find the roots of the quadratic equations by factorisation: x−3x=2(x≠0) View Solution Find the roots of the following equations: x−1x=3,x≠0 View Solution Find the roots of the equationx+1x=3,x≠0 View Solution Find the roots of the following equations: ...
相关知识点: 试题来源: 解析 N+1=cos(2n)/6π+2sin(2n)/6π, 3.o.s =0,-1/2±1/2(√3,-3/2±1/2i W2Uses the 6th roots of unity.Al for 3 correct roots.A1 Gives all six roots of the equation inexact form. 反馈 收藏
Find all roots of the equation,including complex roots.Bingpei Wu
If a,b,c∈R, then find the number of real roots of the equation Δ=∣∣∣∣xc−b−cxab−ax∣∣∣∣=0 View Solution If a,b,c are three real numbers, then number of real roots of ∣∣∣∣xc−b−cxab−ax∣∣∣∣=0 is View Solution ...