ax^2+bx+c=0 ⇔ x=- b±sqrt(b^2-4ac)/2aSince the given equation is already in standard form, we can identify the values of a, b, and c.
6x^2-x-5=0 ⇔ 6x^2+( - 1)x+( - 5)=0
Now, let's evaluate the discriminant.
b We want to use the discriminant of the given quadratic equation to determine the number and type of the roots. If we do not want to know the exact values of the roots, we only need to work with the discriminant. From Part A, we know that the discriminant of the given equation is 121.
Equation:& 6x^2-x-5=0
Discriminant:& 121
Since the discriminant is greater than zero and a perfect square, the quadratic equation has two rational roots.
c We will use the Quadratic Formula to find the exact solutions of the given equation.
x=- b±sqrt(b^2-4 a c)/2 a
Recall that we have already identified the values of a, b, and c in Part A, as well as the discriminant, b^2-4ac.
a= 6, b= - 1, c= - 5
Discriminant: 121
Let's substitute these values into the Quadratic Formula.
