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Grade: 11 

                        
Find the real values of x which satisfy X^2-3x+2>0 and x^2-3x-4= or
2 years ago

Answers : (1)

Susmita
425 Points 
							
Given 
x2-3x-4=0
Or,x2-4x+x-4=0
Or,x(x-4)+1(x-4)=0
Or,(x-4)(x+1)=0
So either x=4 or x=-1.
For x=4,
x2-3x+2=16-12+2=6>0
For 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.
2 years ago
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