Is Y=2x^2+3x-2 real roots?
Yes, the quadratic equation Y=2x^2+3x-2 has real roots. To find the roots, we can use the quadratic formula:
x = (-b ± sqrt(b^2-4ac)) / 2a
In this case, a=2, b=3, and c=-2. Substituting these values into the quadratic formula, we get:
x = (-3 ± sqrt(3^2 - 4(2)(-2))) / 2(2)
x = (-3 ± sqrt(25)) / 4
x = (-3 ± 5) / 4
So the roots are:
x = (-3 + 5) / 4 = 1/2
x = (-3 - 5) / 4 = -2
Both of these roots are real numbers, so the answer is yes.
2x^2+3x-2 = (x+2)(2x-1)
so yes, it has roots of -2 and 1/2
You are correct. We can factor the quadratic equation Y=2x^2+3x-2 as (x+2)(2x-1), which means the roots are -2 and 1/2. Both of these roots are real numbers, so the answer is still yes.
To determine if the quadratic equation Y=2x^2+3x-2 has real roots, we can examine the discriminant of the equation.
The discriminant (denoted as "D") is calculated using the formula D = b^2 - 4ac. In this equation, a, b, and c represent the coefficients from the quadratic equation in the form of ax^2 + bx + c = 0.
In our case, a = 2, b = 3, and c = -2. Substituting these values into the formula, we get:
D = (3)^2 - 4(2)(-2)
D = 9 + 16
D = 25
Since the discriminant is positive (D > 0), the quadratic equation has real roots.