標題:
A. Maths中, Sum and Product of Roots of Quadratic Equation問題
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發問:
以下是問題: Given that α and β are the roots of the quadratic equation x^2 - (k + 1)x + 2k = 0. If (α + 1)(β + 4) = 35, show that α = -k + 10. Hence, find the values of k. 請幫忙一下~~
最佳解答:
As α and β are the real roots of the equation x2 - (k+1)x + 2k = 0, Sum of roots = -[-(k+1)]/1 α+β = (k+1) ................... (1) Product of roots = 2k/1 αβ = 2k ....................... (2) (α+1)(β+4)=35 αβ + 4α + β + 4 = 35 αβ + 3α + (α + β) = 31 2k + 3α + (k+1) = 31 ....... by (1) and (2) 3k + 3α = 30 3α = 30 - 3k α = 10 - k α = -k + 10 So α = -k + 10 Substitute α = -k+10 into (1), -k+10 + β = k+1 β = 2k-9 Substitute α = -k+10 and β = 2k-9 into (2), (-k+10)(2k-9) = 2k -2k2 + 9k + 20k - 90 = 2k -2k2 + 27k - 90 = 0 2k2 - 27k + 90 = 0 (2k-15)(k-6) = 0 2k-15 = 0 or k-6 = 0 k = 15/2 or k = 6 So k = 15/2 or k = 6.
其他解答:
x^2-(k+1)x+2k=0 Therefore α+β=-[-(k+1)/1]=k+1 αβ=2k/1=2k (α+1)(β+4)=35 αβ+4α+β+4=35 2k+k+1+3α=31 3k+3α=30 k+α=10 α=-k+10 Sub α=-k+10 into x^2-(k+1)x+2k=0 (-k+10)^2-(k+1)(-k+10)+2k=0 k^2-20k+100+k^2-9k-10+2k=0 2k^2-27k+90=0 (k-15)(k-6)=90 k=15 or k=6
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