This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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1. Tatua mlinganyo wa quadratic a) X^2 + 25 = 0 Step 1: Tenga X^2 upande mmoja wa mlinganyo. X^2 = -25 Step 2: Tafuta mzizi wa mraba wa pande zote mbili. Kumbuka kuwa sqrt(-1) = j. X = ± sqrt(-25) X = ± sqrt(25 × -1) X = ± sqrt(25) × sqrt(-1) X = ± 5j Suluhisho ni X = ± 5j. b) 4t^2 - 5t + 7 = 0 Huu ni mlinganyo wa quadratic wa fomu at^2 + bt + c = 0, ambapo a=4, b=-5, na c=7. Step 1: Tumia fomula ya quadratic: t = -b ± sqrt(b^2 - 4ac)2a. Step 2: Badilisha thamani za a, b, c kwenye fomula. t = -(-5) ± sqrt((-5)^2 - 4(4)(7))2(4) Step 3: Rahisisha usemi chini ya mzizi wa mraba. t = 5 ± sqrt(25 - 112)8 t = 5 ± sqrt(-87)8 Step 4: Andika sqrt(-87) kama jsqrt(87) (kwa kuwa j = sqrt(-1)). t = 5 ± jsqrt(87)8 Suluhisho ni t = 5 ± jsqrt(87)8. 2. Tathmini a) (-4)/(sqrt(9)) Step 1: Tafuta mzizi wa mraba wa 9. sqrt(9) = 3 Step 2: Gawanya -4 kwa 3. (-4)/(3) Thamani ni -(4)/(3). b) j ( (1+j3)/(1-j2) ) Step 1: Kwanza, rahisha sehemu ya namba changamano kwa kuzidisha nambari na denominata kwa conjugate ya denominata. Conjugate ya 1-j2 ni 1+j2. (1+j3)/(1-j2) × (1+j2)/(1+j2) Step 2: Panua nambari na denominata. Kumbuka kuwa j^2 = -1. Nambari: (1+j3)(1+j2) = 1 + j2 + j3 + j^26 = 1 + j5 - 6 = -5 + j5 Denominata: (1-j2)(1+j2) = 1^2 - (j2)^2 = 1 - j^24 = 1 - (-1)4 = 1 + 4 = 5 Step 3: Badilisha nambari na denominata zilizorahisishwa kwenye sehemu. (-5 + j5)/(5) = (-5)/(5) + (j5)/(5) = -1 + j Step 4: Sasa zidisha matokeo haya kwa j. j(-1 + j) = -j + j^2 = -j - 1 Thamani ni -1 - j. That's 2 down. 3 left today — send the next one.