Question

二次函数求解二次函数求解二次函数求解

Answer 1

√x＞ax+3/2;

x＞a²x²+9/4+3ax;

a²x²+(3a-1)x+9/4＜0;

∴4+b=-(3a-1)/a²;

4b=9/4/a²;

b=9/16a²;

4+9/16a²=-3/a+1/a²

7/16a²-3/a-4=0;

7/a²-48/a-64=0;

(7/a+8)(1/a-8)=0;

7/a+8=0或1/a-8=0;

∴a=-7/8;或a=1/8;

b=36/49或b=36;

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