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A maths問題
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a+b=-3ab=-11.a^3-b^3--------------------------------------------------------------------------------------------------------------------------------2. If a and b are the roots of the quadratic equation 2x^2+ax+b=0,where b≠0,find the values of a and b3. If the equations x^24x+k=0 and 2x^2-3x+k=0 have a common... 顯示更多 a+b=-3 ab=-1 1.a^3-b^3 -------------------------------------------------------------------------------------------------------------------------------- 2. If a and b are the roots of the quadratic equation 2x^2+ax+b=0,where b≠0,find the values of a and b 3. If the equations x^24x+k=0 and 2x^2-3x+k=0 have a common root a,find the values of k
(1) a3-b3 =(a-b)(a2+ab+b2) =(a-b)[(a+b)2-ab] =(a-b)[(-3)2-(-1)] =10(a-b) (2) 2x2+ax+b=0 a+b=-a/2---(1) ab=b/2---(2) (1)=>3a/2+b=0 b=-3a/2---(3) Sub (3) into (2) a(-3a/2)=(-3a/2)/2 -3a2/2=-3a/4 6a2=3a 2a2-a=0 a(a-1)=0 a=0 or a=1 When a=0, (1)=>b=0 Since b≠0, a≠0 Therefore a=1---(4) Sub (4) into (3): b=-3(1)/2 b=-3/2 Therefore a=1,b=-3/2 (3) Let the common root be a a2+24a+k=2a2-3a+k a2-2a2+24a+3a=0 -a2+27a=0 a(a-27)=0 a=0 or a=27 From the equation x2+4x+k=0 When a=0,k=0; When a=27,k=-810 Therefore, k=0 or k=-810
其他解答:
A maths問題
發問:
a+b=-3ab=-11.a^3-b^3--------------------------------------------------------------------------------------------------------------------------------2. If a and b are the roots of the quadratic equation 2x^2+ax+b=0,where b≠0,find the values of a and b3. If the equations x^24x+k=0 and 2x^2-3x+k=0 have a common... 顯示更多 a+b=-3 ab=-1 1.a^3-b^3 -------------------------------------------------------------------------------------------------------------------------------- 2. If a and b are the roots of the quadratic equation 2x^2+ax+b=0,where b≠0,find the values of a and b 3. If the equations x^24x+k=0 and 2x^2-3x+k=0 have a common root a,find the values of k
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最佳解答:(1) a3-b3 =(a-b)(a2+ab+b2) =(a-b)[(a+b)2-ab] =(a-b)[(-3)2-(-1)] =10(a-b) (2) 2x2+ax+b=0 a+b=-a/2---(1) ab=b/2---(2) (1)=>3a/2+b=0 b=-3a/2---(3) Sub (3) into (2) a(-3a/2)=(-3a/2)/2 -3a2/2=-3a/4 6a2=3a 2a2-a=0 a(a-1)=0 a=0 or a=1 When a=0, (1)=>b=0 Since b≠0, a≠0 Therefore a=1---(4) Sub (4) into (3): b=-3(1)/2 b=-3/2 Therefore a=1,b=-3/2 (3) Let the common root be a a2+24a+k=2a2-3a+k a2-2a2+24a+3a=0 -a2+27a=0 a(a-27)=0 a=0 or a=27 From the equation x2+4x+k=0 When a=0,k=0; When a=27,k=-810 Therefore, k=0 or k=-810
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