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 real roots of equation x^3 - 3 x^2 - x = -2. Find the real roots of the equation. x^2 - 2 x + 2 = 0 Find the real roots of the equation. 2 x^2 - 5 x - 3 = 0 Explore our homework questions and answers library ...
Find the sum of all the real roots of the equation P(x)= 2x^98+3x^97+2x^96+…….+2x+3=0.
Find all roots of the equation,including complex roots.Bingpei Wu
【题目】Example 9: Find the sum of the real roots of the equation$$ x ^ { 2 } + 1 2 x + 1 6 = 2 \sqrt { x ^ { 2 } + 1 2 x + 1 9 } $$ 相关知识点: 试题来源: 解析 【解析】 Solution:-12. Method 1: Rewrite the original equation as $$ x ^ { 2 } + 1 2...
Example 9: Find the sum of the real roots of the equation x^2+12x+16=2√(x^2+12x+19) 相关知识点: 试题来源: 解析 Solution:-12. Method 1: Rewrite the original equation as x^2+12x+16-2√(x^2+12x+19)=0 Or (√(x^2+12x+19))^2-2√(x^2+12x+19)-3=0 (√(x^2+...
View Solution Find the roots of the equationx+1x=3,x≠0 View Solution Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class 12, IIT JEE prep, NEET preparation and CBSE, UP Board, Bihar...
Answer to: Find the number of roots of the equation cot x = pi/2 + x in -pi, 3 pi/2. By signing up, you'll get thousands of step-by-step solutions...
Answer to: Find the five roots of the equation (x+1)^5=32x^5 By signing up, you'll get thousands of step-by-step solutions to your homework...