Consider the following quadratic equation:
2y^
2−1=5y
Step 2 of 2 : Use the discriminant, b2−4ac
, to determine the number of solutions of the given quadratic equation. Then solve the quadratic equation using the formula y=−b±b2−4ac⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯/2a
.
To determine the number of solutions of the given quadratic equation, we need to calculate the discriminant, which is given by b^2 - 4ac.
In this equation, the coefficient of y^2 is 2, the coefficient of y is -1, and the constant term is -5.
So, the discriminant is:
b^2 - 4ac = (-1)^2 - 4(2)(-5) = 1 + 40 = 41
The discriminant is positive, specifically 41.
Since the discriminant is positive, there are two distinct solutions to the quadratic equation.
Now, to solve the quadratic equation using the formula:
y = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values:
y = (-(-1) ± √(1 - 4(2)(-5))) / (2(2))
y = (1 ± √(1 + 40)) / 4
y = (1 ± √41) / 4
So, the two solutions to the quadratic equation are:
y = (1 + √41) / 4
y = (1 - √41) / 4