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(b) Los op vir \( x \) : (1) \( \quad 2 a x+b=c x \) (3) \( a x^{2}-9 b^{2}=0 \) (5) \( \quad x^{2}+3 b x+2 b^{2}=0 \) (7) \( \quad x^{2}-2 b x+a x-2 a b=0 \)

Ask by Mann Li. in South Africa
Mar 16,2025

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Solutions for the equations are: 1. \( x = \frac{-b}{2a - c} \) 2. \( x = \pm \frac{3b}{\sqrt{a}} \) 3. \( x = -b \) or \( x = -2b \) 4. \( x = -a \) or \( x = 2b \)

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Om die vergelykings op te los, kan ons verskeie algebraïese metodes gebruik: (1) Voor die eerste vergelyking \( 2ax + b = cx \) kan ons dit herorganiseer om \( x \) te isolateer: \( 2ax - cx = -b \) wat lei tot \( x(2a - c) = -b \) en dan \( x = \frac{-b}{2a - c} \). (3) Die tweede vergelyking \( ax^{2} - 9b^{2} = 0 \) is 'n kwadratiese vergelyking. Ons kan dit herskryf as \( ax^{2} = 9b^{2} \) en \( x^{2} = \frac{9b^{2}}{a} \). Neem die vierkantswortel om \( x \) te kry: \( x = \pm \frac{3b}{\sqrt{a}} \). (5) Vir \( x^{2} + 3bx + 2b^{2} = 0 \) kan ons die kwadratiese formule gebruik: \( x = \frac{-3b \pm \sqrt{(3b)^{2} - 4 \cdot 1 \cdot 2b^{2}}}{2 \cdot 1} \), wat vereenvoudig tot \( x = \frac{-3b \pm \sqrt{b^{2}}}{2} \) en uiteindelik \( x = -b \) of \( x = -2b \). (7) Vir \( x^{2} - 2bx + ax - 2ab = 0 \) kan ons dit reorganiseer as \( x^{2} + (a - 2b)x - 2ab = 0 \). Gebruik die kwadratiese formule om \( x \) te vind: \( x = \frac{-(a - 2b) \pm \sqrt{(a - 2b)^{2} + 8ab}}{2} \). Probeer om die waardes in nie te vinnig te veralgemeen nie; neem jou tyd om elke stap te verstaan.

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