Given $2x^2-1=2x$ , how do you find the discriminant and the number of solutions?

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Answer 1 · 2018-03-29T12:04:51

Solution: $x = 1/2+sqrt3/2 and x = 1/2-sqrt3/2$

$2x^2-1=2x or 2x^2-2x-1=0$ Comparing with standard

quadratic equation $ax^2+bx+c=0$ we get,

$a=2 ,b=-2 ,c=-1$ Discriminant $D= b^2-4ac$ or

$D=4+8 =12$ If discriminant positive, we get two real

solutions, if it is zero we get just one solution, and if it is negative

we get complex solutions.Discriminant is positive here , so it has

two real roots . Quadratic formula: $x= (-b+-sqrtD)/(2a)$ or

$x= (2+-sqrt12)/4 = (2+-2sqrt3)/4= 1/2+-sqrt3/2$

Solution: $x = 1/2+sqrt3/2 and x = 1/2-sqrt3/2$ [Ans]

Given 2x^2-1=2x2x^2-1=2x, how do you find the discriminant and the number of solutions?