Leta,bandcbe three real and distinct numbers. If the difference of the roots of the equationax2+bx−c=0is 1, then roots of the equationax2−ax+c=0are 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)...
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...
If ? and B are the roots of the equation,$$ 7 x ^ { 2 } - 3 x - 2 = 0 $$D, then the value of$$ \frac { \alpha } { 1 - \alpha ^ { 2 } } + \frac { \beta } { 1 - \beta ^ { 2 } } $$is equal to:C A $$ \frac { 1 } { 2 4 } $$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. ...
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...
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
If α and β are the roots of the equations x2−2x−1=0, then what is the value of α2β−2+β2α−2 A−2 B0 C30 D34Submit If αandβ are the roots of the equation 2x2−3x+4=0, then α2+β2 = ___ A14 B74 C−74 D−14Submit If αandβ are roots of...
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 ...
几道英文题!第37页1.If the equation(方程) m(x-1)=2001-n(x-2) for x has infinite roots(无限个根) ,then m的2001次方+n的2001次方=2.We have the following numbers 9/5,12/7,27/17,36/19,54/29,the maximum (最大的)number