what is the value of the discriminant and of the real solutions of the quadratic equation?
5x^2-9x plus 2 equals 0
why not just use conventional math notation?
5x^2-9x+2 = 0
the discriminant is b^2-4ac = 81-40 = 41
Now just use that in the quadratic formula. Since the discriminant is positive, there are two real solutions:
x = (9±√41)/10
To find the value of the discriminant and the real solutions of a quadratic equation, we need to use the quadratic formula.
The quadratic formula is given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In the quadratic equation 5x^2 - 9x + 2 = 0, we can identify the values of a, b, and c:
a = 5
b = -9
c = 2
Now we can substitute these values into the quadratic formula to find the solutions.
Step 1: Calculate the discriminant.
The discriminant is given by Δ = b^2 - 4ac.
In this case, the discriminant is Δ = (-9)^2 - 4(5)(2).
Calculating:
Δ = 81 - 40
Δ = 41
So, the value of the discriminant is 41.
Step 2: Calculate the real solutions using the quadratic formula.
Using the formula x = (-b ± √(b^2 - 4ac)) / (2a), we can substitute the values of a, b, c, and the discriminant Δ into the formula.
x = (-(-9) ± √((9)^2 - 4(5)(2))) / (2(5))
x = (9 ± √(81 - 40)) / 10
x = (9 ± √41) / 10
The two solutions are:
x₁ = (9 + √41) / 10
x₂ = (9 - √41) / 10
Therefore, the real solutions of the quadratic equation 5x^2 - 9x + 2 = 0 are:
x₁ = (9 + √41) / 10
and
x₂ = (9 - √41) / 10