For what values of m will x^2-2/2x-3 = m-1/m+1 havea.real roots equal in magnitude but opposite in sign?m=1)b.two real,non-equal roots?m 大于 -3,m 不等于 1）
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(x^2-2)/(2x-3)=(m-1)/(m+1)
(x^2-2)(m+1)=(m-1)(2x-3)
(m+1)x^2-2(m+1)=2(m-1)x-3(m-1)
(m+1)x^2-2(m-1)x+3m-3-2m-2=0
(m+1)x^2-2(m-1)x+m-5=0
a.real roots equal in magnitude but opposite in sign
2(m-1)/(m+1)=0
m=1b.two real,non-equal roots
△=[-2(m-1)]^2-4(m+1)(m-5)>0m^2-2m+1-m^2+4m+5>02m+6>0m>-3and m+10,m-1
so m>-3 and m-1
======以下答案可供参考======
供参考答案1:
x^2 -2/2x - 3 不就是 x^2 - x - 3 么。你确定没打错？
供参考答案2:
第一题存在绝对值大小相等但符号相反的实数根
x²-2x(m-1)/(m+1)+(m-5)/(m+1)=0
(m-1)/(m+1)=0
m=1第二题存在不等的实数根
△=(2m+6)/(m+1)>0解得m-1供参考答案3:
Let x1,x2 be roots of this equtaion x^2-2/2x-3 = m-1/m+1,
when x1,x2 are not equal 3/2,and m is not equal -1,
then (m+1)x^2-2(m-1)x+(m-5) = 0
(1)becuse real roots equal in magnitude but opposite in sign of this equation, so
Vieta's Theorem (韦达定理)
x1+ x2 = -b/a = 2(m-1)/(m+1) = 0 then m = 1
(2) becuse two real, non-equal roots of this equation , only △=b^2-4ac >0i.e △=4*(m-1)^2-4(m+1)(m-5) >0, Reduction to m+3>0 then m>-3 In conclusion, the Scope of m is m >-3 and m not equal 1.