已知一元二次方程a(b-c)x*2+b(c-a)x+c(a-b)=0有两个相等的实根,a,b,c是三角形ABC的三边,且满足2b=a+c.(1)求a:b:c(2)若上述三角形最短边为5,而方程x^2-(m+2)x+m-3的两根平方和为最长边的3倍,求m的值
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![已知一元二次方程a(b-c)x*2+b(c-a)x+c(a-b)=0有两个相等的实根,a,b,c是三角形ABC的三边,且满足2b=a+c.(1)求a:b:c(2)若上述三角形最短边为5,而方程x^2-(m+2)x+m-3的两根平方和为最长边的3倍,求m的值](/uploads/image/z/316717-61-7.jpg?t=%E5%B7%B2%E7%9F%A5%E4%B8%80%E5%85%83%E4%BA%8C%E6%AC%A1%E6%96%B9%E7%A8%8Ba%28b-c%29x%2A2%2Bb%28c-a%29x%2Bc%28a-b%29%3D0%E6%9C%89%E4%B8%A4%E4%B8%AA%E7%9B%B8%E7%AD%89%E7%9A%84%E5%AE%9E%E6%A0%B9%2Ca%2Cb%2Cc%E6%98%AF%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E7%9A%84%E4%B8%89%E8%BE%B9%2C%E4%B8%94%E6%BB%A1%E8%B6%B32b%3Da%2Bc.%EF%BC%881%EF%BC%89%E6%B1%82a%EF%BC%9Ab%EF%BC%9Ac%EF%BC%882%EF%BC%89%E8%8B%A5%E4%B8%8A%E8%BF%B0%E4%B8%89%E8%A7%92%E5%BD%A2%E6%9C%80%E7%9F%AD%E8%BE%B9%E4%B8%BA5%2C%E8%80%8C%E6%96%B9%E7%A8%8Bx%5E2-%28m%2B2%29x%2Bm-3%E7%9A%84%E4%B8%A4%E6%A0%B9%E5%B9%B3%E6%96%B9%E5%92%8C%E4%B8%BA%E6%9C%80%E9%95%BF%E8%BE%B9%E7%9A%843%E5%80%8D%2C%E6%B1%82m%E7%9A%84%E5%80%BC)
已知一元二次方程a(b-c)x*2+b(c-a)x+c(a-b)=0有两个相等的实根,a,b,c是三角形ABC的三边,且满足2b=a+c.(1)求a:b:c(2)若上述三角形最短边为5,而方程x^2-(m+2)x+m-3的两根平方和为最长边的3倍,求m的值
已知一元二次方程a(b-c)x*2+b(c-a)x+c(a-b)=0有两个相等的实根,a,b,c是三角形ABC的三边,且满足
2b=a+c.(1)求a:b:c(2)若上述三角形最短边为5,而方程x^2-(m+2)x+m-3的两根平方和为最长边的3倍,求m的值.
已知一元二次方程a(b-c)x*2+b(c-a)x+c(a-b)=0有两个相等的实根,a,b,c是三角形ABC的三边,且满足2b=a+c.(1)求a:b:c(2)若上述三角形最短边为5,而方程x^2-(m+2)x+m-3的两根平方和为最长边的3倍,求m的值
1
一元二次方程a(b-c)x*2+b(c-a)x+c(a-b)=0有两个相等的实根
那么判别式
Δ=b²(c-a)²-4ac(a-b)(b-c)=0
∴b²(c²-2ca+a²)-4ac(ab-b²-ac+bc)
=b²c²-2ab²c+a²b²-4a²bc+4ab²c+4a²c²-4abc²
=a²b²+b²c²+4a²c²+2ab²c-4a²bc-4abc²
=(ab+bc-2ac)²=0
∴ab+bc-2ac=0
∵2b=a+c
∴b=(a+c)/2
∴(a+c)²/2-2ac=0
∴(a-c)²=0
∴a=c=b
2
a=b=c=5
方程x^2-(m+2)x+m-3=0的两根为x1,x2
Δ=(m+2)²-4(m-3)≥0恒成立
则x1+x2=m+2,x1x2=m-3
∴x²1+x²2
=(x1+x2)²-2x1x2
=(m+2)²-2(m-3)
=m²+2m+10
根据题意:
m²+2m+10=15
∴m²+2m-5=0
解得:m=-1±√6
有问题再追问吧