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唔識做功課34
就快要將標題改為(做極都唔識)>0<巳經操練2星期仍是這樣!!!1. If a and b are positive constants and a^2 > 4b , which of the following equations must have real roots with the same sign?A. x^2+ax-b=0 B. x^2-ax+b=0 C. x^2-ax-b=02. If m and n are positive constants, which of the following equations must have real roots... 顯示更多 就快要將標題改為(做極都唔識)>0<巳經操練2星期仍是這樣!!! 1. If a and b are positive constants and a^2 > 4b , which of the following equations must have real roots with the same sign? A. x^2+ax-b=0 B. x^2-ax+b=0 C. x^2-ax-b=0 2. If m and n are positive constants, which of the following equations must have real roots with different signs? A. x^2+mx+n=0 B.3x^2+mx-n=0 C. -x^2+mx-n=0
最佳解答:
1. The answer : (B) x2 - ax + b = 0 A. false x2 + ax - b = 0, where a and b are positive The discriminant, Δ = a2 - 4b > 0 for a2 >4b There are two real roots. Factorize x2 + ax - b into the form of (x + h)(x - k) whereh and k are positive Then, (x + h)(x - k) = 0 x = -h or x = k The two roots have different signs. B. true x2 - ax + b = 0 The discriminant, Δ = (-a)2 - 4b = a2 -4b > 0 for a2 > 4b There are two real roots. Factorize x2 - ax + b into the form of (x - h)(x - k) whereh and k are positive Then, (x - h)(x - k) = 0 x = h or x = k The two roots have the same sign. C. false x2 - ax - b = 0, where a and b are positive The discriminant, Δ = (-a)2 - 4(-b) = a2 +4b > 0 for a2 > 4b There are two real roots. Factorize x2 - ax - b into the form of (x + h)(x - k) whereh and k are positive Then, (x + h)(x - k) = 0 x = -h or x = k The two roots have different signs. ==== 2. The answer : (B) 3x2 + mx - n = 0 A. false x2 + mx + n = 0, where m and n are positive The discriminant, Δ = m2 - 4n There is no sufficient information to determine whether Δ is positive ornegative. The equation may not have real roots.(if Δ is negative) B. true 3x2 + mx - n = 0, where m and n are positive The discriminant, Δ = m2 - 4(3)(-n) = m2 +12n > 0 for m and n are positive There are two real roots. Factorize 3x2 + mx - n into the form of (3x + h)(x - k) or (3x- h)(x + k) where h and k are positive Then, (3x + h)(x - k) = 0 or (3x - h)(x + k) = 0 (x = -h/3 or x = k) or (x = h/3 or x = -k) The two roots have different signs. C. false -x2 + mx - n = 0, where m and n are positive The discriminant, Δ = m2 - 4(-1)(-n) = m2 - 4n There is no sufficient information to determine whether Δ is positive ornegative. The equation may not have real roots.(if Δ is negative)
其他解答:
^0^測驗有90分呀!多謝你嘅鼓勵與教導|||||yan妹, 或者你試下重做一D已經做過的題目, 思考一下當中的邏輯。 中學數好的地方係, 你不斷 train 總會學到~ 因為來去都係果D野, 變化唔多...
唔識做功課34
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發問:就快要將標題改為(做極都唔識)>0<巳經操練2星期仍是這樣!!!1. If a and b are positive constants and a^2 > 4b , which of the following equations must have real roots with the same sign?A. x^2+ax-b=0 B. x^2-ax+b=0 C. x^2-ax-b=02. If m and n are positive constants, which of the following equations must have real roots... 顯示更多 就快要將標題改為(做極都唔識)>0<巳經操練2星期仍是這樣!!! 1. If a and b are positive constants and a^2 > 4b , which of the following equations must have real roots with the same sign? A. x^2+ax-b=0 B. x^2-ax+b=0 C. x^2-ax-b=0 2. If m and n are positive constants, which of the following equations must have real roots with different signs? A. x^2+mx+n=0 B.3x^2+mx-n=0 C. -x^2+mx-n=0
最佳解答:
1. The answer : (B) x2 - ax + b = 0 A. false x2 + ax - b = 0, where a and b are positive The discriminant, Δ = a2 - 4b > 0 for a2 >4b There are two real roots. Factorize x2 + ax - b into the form of (x + h)(x - k) whereh and k are positive Then, (x + h)(x - k) = 0 x = -h or x = k The two roots have different signs. B. true x2 - ax + b = 0 The discriminant, Δ = (-a)2 - 4b = a2 -4b > 0 for a2 > 4b There are two real roots. Factorize x2 - ax + b into the form of (x - h)(x - k) whereh and k are positive Then, (x - h)(x - k) = 0 x = h or x = k The two roots have the same sign. C. false x2 - ax - b = 0, where a and b are positive The discriminant, Δ = (-a)2 - 4(-b) = a2 +4b > 0 for a2 > 4b There are two real roots. Factorize x2 - ax - b into the form of (x + h)(x - k) whereh and k are positive Then, (x + h)(x - k) = 0 x = -h or x = k The two roots have different signs. ==== 2. The answer : (B) 3x2 + mx - n = 0 A. false x2 + mx + n = 0, where m and n are positive The discriminant, Δ = m2 - 4n There is no sufficient information to determine whether Δ is positive ornegative. The equation may not have real roots.(if Δ is negative) B. true 3x2 + mx - n = 0, where m and n are positive The discriminant, Δ = m2 - 4(3)(-n) = m2 +12n > 0 for m and n are positive There are two real roots. Factorize 3x2 + mx - n into the form of (3x + h)(x - k) or (3x- h)(x + k) where h and k are positive Then, (3x + h)(x - k) = 0 or (3x - h)(x + k) = 0 (x = -h/3 or x = k) or (x = h/3 or x = -k) The two roots have different signs. C. false -x2 + mx - n = 0, where m and n are positive The discriminant, Δ = m2 - 4(-1)(-n) = m2 - 4n There is no sufficient information to determine whether Δ is positive ornegative. The equation may not have real roots.(if Δ is negative)
其他解答:
^0^測驗有90分呀!多謝你嘅鼓勵與教導|||||yan妹, 或者你試下重做一D已經做過的題目, 思考一下當中的邏輯。 中學數好的地方係, 你不斷 train 總會學到~ 因為來去都係果D野, 變化唔多...
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