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Quadratic equation mixed asked by ربيع عامود in the نادي الرياضيات


Exercise:

Solve in R , (x25x+4)23(x25x+4)+1=x

Solution: Let u(x)=x25x+4

So u(x)=x24xx+4=x(x1)4(x1)=(x1)(x4)

Thus (x1)2(x4)23(x4)(x1)(x1)=0

(x1)[(x1)(x4)23(x4)1]=0

x=1or(x1)(x4)23(x4)1=0

x(x4)2(x4)23(x4)1=0(x4)(x(x4)3)(x4)21=0

(x4)(x24x3)(x4)21=0(x4)[x24x3x+4]1=0

(x4)(x25x+1)1=0(x4)(x(x5)+1)1=0

 x(x5)(x4)+x41=0(x5)(x(x4)+1)=0

So x=5orx24x+44+1=0(x2)23=0x=2±3

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