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
.
The given quadratic equation is 2y^2 − 1 = 5y.
To determine the number of solutions of the quadratic equation, we can use the discriminant, which is given by b^2 − 4ac.
Comparing the equation to the standard form of a quadratic equation, ax^2 + bx + c = 0, we have:
a = 2, b = -5, and c = -1.
The discriminant is b^2 − 4ac:
Discriminant = (-5)^2 − 4(2)(-1)
= 25 + 8
= 33
Since the discriminant is positive (33 > 0), the quadratic equation has two real solutions.
Now, let's solve the quadratic equation using the quadratic formula:
y = (−b ± √(b^2 − 4ac)) / (2a)
Plugging in the values we have:
y = (−(-5) ± √(33)) / (2(2))
= (5 ± √(33)) / 4
Therefore, the solutions to the given quadratic equation are y = (5 + √(33)) / 4 and y = (5 - √(33)) / 4.