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標題:
a-maths難題
發問:
prove that, for all real values of a , the equation x^2+2ax+2a^2+a+1=0 has no real roots for x 有有冇人可以解答用解得詳細d,,
最佳解答:
prove that, for all real values of a , the equation x2+2ax+2a2+a+1=0 has no real roots for x x2+2ax+2a2+a+1=0 則判別式 B2 – 4AC = (-2a)2 – 4(1)(2a2+a+1) = 4a2 – 8a2 – 4a – 4 = -4a2 – 4a – 4 = -4(a2 + a) – 4 = -4(a2 + a + 0.52 – 0.52) – 4 = -4[(a + 0.5)2 – 0.52] – 4 = -4(a + 0.5)2 + 4*0.52 – 4 = -4(a + 0.5)2 – 3 所以當無論 a 為何值， 判別式必小於零，所以這方程式沒有實根。
prove that, for all real values of a , the equation x^2+2ax+2a^2+a+1=0 has no real roots for x Put x^2+2ax+2a^2+a+1=0 into B^2-4AC. B^2-4AC =(2a)^2-(4)(1)(2a^2+a+1) =4a^2-8a^2-4a-4 =-4a^2-4a-4 =-4a^2-4a+8-12 =-4(a^2+a-2)-12 =-4(a+2)(a-1)-12 The largest value of -4(a+2)(a-1)-12 is, -4[(-0.5)+2][(-0.5)-1)-12=(-4)(-1.5)(1.5)-12=9-12=-3 So, the equation x^2+2ax+2a^2+a+1=0 has no real roots for x
a-maths難題
發問:
prove that, for all real values of a , the equation x^2+2ax+2a^2+a+1=0 has no real roots for x 有有冇人可以解答用解得詳細d,,
最佳解答:
prove that, for all real values of a , the equation x2+2ax+2a2+a+1=0 has no real roots for x x2+2ax+2a2+a+1=0 則判別式 B2 – 4AC = (-2a)2 – 4(1)(2a2+a+1) = 4a2 – 8a2 – 4a – 4 = -4a2 – 4a – 4 = -4(a2 + a) – 4 = -4(a2 + a + 0.52 – 0.52) – 4 = -4[(a + 0.5)2 – 0.52] – 4 = -4(a + 0.5)2 + 4*0.52 – 4 = -4(a + 0.5)2 – 3 所以當無論 a 為何值， 判別式必小於零，所以這方程式沒有實根。
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其他解答:prove that, for all real values of a , the equation x^2+2ax+2a^2+a+1=0 has no real roots for x Put x^2+2ax+2a^2+a+1=0 into B^2-4AC. B^2-4AC =(2a)^2-(4)(1)(2a^2+a+1) =4a^2-8a^2-4a-4 =-4a^2-4a-4 =-4a^2-4a+8-12 =-4(a^2+a-2)-12 =-4(a+2)(a-1)-12 The largest value of -4(a+2)(a-1)-12 is, -4[(-0.5)+2][(-0.5)-1)-12=(-4)(-1.5)(1.5)-12=9-12=-3 So, the equation x^2+2ax+2a^2+a+1=0 has no real roots for x
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