use dicriminant to determine whether can be factored and factor if not prime 2x^2+3x-5?

Answer 1

Okay. Do you know the quadratic equation? It is:
x= [-b +/- (square root of b^2 - 4ac)]/2a.

the part that says: (square root of b^2 - 4ac) is the part that is the disciminant. okay so the equation is 2x^2 + 3x - 5. The numbers are the letters. So A=2, B=3, and C=-5. So plug in for the numbers into (square root of b^2 - 4ac). You get. (3^2 - 4 X 2 X -5). That equals (9 - -40) which equals (49). So that is the square root of 49. Whatever is under the radical in a quadratic equation is the dicriminant. If the number is positive, you most it can be factored. If it is a negative it is imaginary. But you know that the square root of 49 is 7 so there for it can be factored. So finish plugging in for factoring. You get: x= [-3 + or - (square root of 49)]/2x2. That equals [-3 + or - (7)]/4. Now break it down into two equations. x=(-3+7)/4 and x=(-3-7)/4. Now solve. You get x=4/4 or 1, and x=-10/4 or -5/2. So therefore your answer will be x=1 and x=-5/2. If you plug the answers in, your equation will come out to be 0.

Answer 2

Let f(x)=2x^2+3x-5=ax^2+bx+c. Then a=2,b=3, c=-5; the discriminant is D = b^2-4ac = 3^2-4*2*-5 = 9+40 = 49. Thus its roots are x1 = (-b + sqrt(D))/(2a) = (-3+7)/(2*2) = (4)/(4) = 1 and x2= (-b - sqrt(D))/(2a) = (-3-7)/(2*2) = -10/4 = -5/2. Thus 2x^2+3x-5 = (x-1)*(2x+5).

Answer 3

D=b^2-4ac=9+40 so can be factored
x1=(-3+7)/4 and x2=(-3-7)/4
therefore the factors are (x-1)(x+5/2)=>(x-1)(2x+5)