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Question

Solve each of the following quadratic equations:
12abx2(9a28b2)x6ab=0

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Solution

(1) Using factorization

12abx2(9a28b2)x6ab=012abx29xa28b2x6ab=03ax(4bx3a)+2b(4bx3a)=0(3ax+2b)(4bx3a)=03ax+2b=0 or 4bx3a=0x=2b3a or 3a4b

(2) Using quadratic formula

12abx2(9a28b2)x6ab=0

Discriminant, D=b24ac=[(9a28b2)]24(12ab)(6ab)=81a4+144a2b2+64b4+288a2b2=81a4+144a2b2+64b4=(9a2+8b2)20

As D0, therefore, the roots are real.

x=b±D2a=(9a28b2)±(9a2+8b2)22(12ab)=(9a28b2)±(9a2+8b2)24ab=(9a28b2)+(9a2+8b2)24ab or (9a28b2)(9a2+8b2)24ab=9a28b2+9a2+8b224ab or 9a28b29a28b224ab=18a224ab or 16b224ab=3a4b or 2b3a


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