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2 solutions
Typo: 2 x 2 − 5 x − 3 = 0
2 x 2 − 5 x 2 x 2 − 5 x − 3 2 x 2 − 6 x + x − 3 2 x ( x − 3 ) + ( x − 3 ) ( 2 x + 1 ) ( x − 3 ) ⟹ x = 2 − 1 , x = 3 = 3 = 0 = 0 = 0 = 0
Hence the roots of this equation are real and distinct.
The nature of the roots can be found out by finding the discriminant which can in turn be found by applying the formula:- b 2 − 4 a c
The equation can be written as 2 x 2 − 5 x − 3 = 0
Here, a = 2 ; b = − 5 ; c = − 3
D = ( − 5 ) 2 − ( 4 × 2 × − 3 )
D = 2 5 − ( − 2 4 )
D = 2 5 + 2 4
D = 4 9 which is greater than zero and a perfect square.
So, the roots of the equation are real, rational and distinct.