Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.
1.6x2 + 2x + 2.5 = 0
I need help
a = 1.6
b = 2
c = 2.5
x = (-b ± √(b^2 - 4ac)/(2a)
(you MUST memorize this formula)
x = (-2 ± (√(4 - 4(1.6)(2.5))/3.2
= (-2 ± √-12)/3.2
This is a complex or imaginary number, so there is no REAL solution.
Is that your problem?
Have you studied complex numbers?
To solve the equation 1.6x^2 + 2x + 2.5 = 0 using the quadratic formula, we need to identify the coefficients of the quadratic terms.
The general form of a quadratic equation is ax^2 + bx + c = 0, where a, b, and c are coefficients.
In our equation, the coefficient of x^2 is 1.6, the coefficient of x is 2, and the constant term is 2.5.
Now, we can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a.
Plugging in the values, we get x = (-(2) ± √((2)^2 - 4(1.6)(2.5))) / (2(1.6)).
Simplifying further, x = (-2 ± √(4 - 20)) / 3.2.
Now, let's calculate the discriminant, which is the expression inside the square root: (4 - 20) = -16.
Since the discriminant is negative, we will have two complex solutions.
x = (-2 ± √(-16)) / 3.2.
To find the square root of a negative number, we can use the imaginary unit i, which is defined as √(-1).
Now, x = (-2 ± 4i) / 3.2.
Simplifying further, x = -2/3.2 ± 1.25i.
Therefore, the solutions to the equation 1.6x^2 + 2x + 2.5 = 0 using the quadratic formula are x = -2/3.2 + 1.25i and x = -2/3.2 - 1.25i.