If the difference of the roots of the equation ax2+bx−c=0 is 1, then roots of the equation ax2−ax+c=0 are always (b≠0) View Solution If a b c are distinct positive real numbers such that b(a+c)=2ac and given equation is ax^(2)+2bx+c=0 then (A) roots of given...
Step by step video & image solution for If the ratio of roots of equation lx^(2) + nx + n= 0 is p: q then find the value sqrt((p)/(q)) + sqrt((q)/(p)) + sqrt((n)/(l))= ? by Maths experts to help you in doubts & scoring excellent marks in Class 10 exams.Updated...
Quadratic Equation(二次方程式的问题)let k be a constant,if A and B are the roots of the equation x2 - 3x +k = 0,the A2 + 3B = ?k 是恒数,如果A和B是x 2- 3x+ k= 0的根,A2 + 3B = ? 相关知识点: 试题来源: 解析 因为A和B是x 2 - 3x+ k= 0的两个根,所以A+B=3,即B...
解析 Solution: Let the roots beand. From Vieta's formulas,From (3),. Using this on (1) and (2) yieldsand. By solving each equation forand equating the result ing expressions, we get. This is equivalent to. Since, and the second factor has negative discriminant, we only have. ...
Suppose that the coefficients of a polynomial equation are Independent random variables defined on subsets of real numbers, The purpose of this paper is to find the exact probability that all roots of a random polynomial equation are real. Since a polynomial equation of degree higher than four ...
If difference of roots of the equation x2−px+q=0 is 1, then p2+4q2 equals- View Solution If p and q are the roots of the quadratic equation x2+px−q=0, then find the values of p and q. View Solution The roots of the equation (p−q)x2+(q−r)x+(r−p)=0 ...
IF the difference between the roots the roots of the equation x^2 +ax +1=0 is less than sqrt(5) then the set of possible values of a is
b are the roots of a quadratic equation$$ x ^ { 2 } - 3 x + 5 = 0 $$then the equation whose roots are$$ \left\{ a ^ { 2 } - 3 a + 7 \right\} a n d ( b ^ { 2 } - 3 b $$B+7)is:$$ x ^ { 2 } + 4 x + 1 = 0 $$10$$ x ^ { 2 } - 4 x...
百度试题 结果1 题目If the roots of a quadratic equation are A + 1 and A - 1, write the equation in ax^2+bx+c=0 form. 相关知识点: 试题来源: 解析 x^2 − 2Ax+A^2 − 1 = 0x^2 − 2Ax+A^2 − 1 = 0 反馈 收藏 ...
展开得mx-m=2001-nx+2n 整理得(m+n)x=2001+2n+m 因为x有infinite roots，即无限解，所以方程与x无关，即m+n=0 所以有m+n=0且2001+2n+m=0 两式联立，得m=2001,n=-2001 所以m^2001+n^2001=2001^2001+(-2001)^2001=0 结果是零咯 wes...