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Find the real values of x which satisfy X^2-3x+2>0 and x^2-3x-4= or
Given x2-3x-4=0Or,x2-4x+x-4=0Or,x(x-4)+1(x-4)=0Or,(x-4)(x+1)=0So either x=4 or x=-1.For x=4,x2-3x+2=16-12+2=6>0For x=-1,x2-3x+2=1+3+2=6>0.So the values are x=4 and x=-1.Please approve the answer if you are helped.
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