What are the real roots of the quadratic function 4x^2-x+5?
If it is 4x^2 - x - 5 = (4x-5)(x+1)
Do you have a typo?
no typo. I tried the quadratic equation and got -.986 and 1.24. Does this mean there are no real roots (an option in the question).
Also, for the question 4x^2-12x+9, is the real root (2x-3)?
solving 4x^2 - x + 5 = 0
x = (1 ± √-79)/8
which is imaginary.
So there are no real roots.
(How did you get -.986 and 1.24 ??)
For 4x^2 - 12x + 9
it factors to (2x-3)^2
the roots of the corresponding equation ....
4x^2 - 12x + 9 = 0
(2x-3)^2 = 0
x = 3/2
So the real root is 3/2
To find the real roots of the quadratic function 4x^2 - x + 5, we can use the quadratic formula.
The quadratic formula states that for a quadratic equation in the form ax^2 + bx + c = 0, the roots can be found using the formula:
x = (-b ± √(b^2 - 4ac)) / 2a.
In this case, a = 4, b = -1, and c = 5. By substituting these values into the quadratic formula, we can find the roots of the quadratic function.
x = (-(-1) ± √((-1)^2 - 4 * 4 * 5)) / (2 * 4)
= (1 ± √(1 - 80)) / 8
= (1 ± √(-79)) / 8.
Since the term inside the square root is negative, √(-79) is not a real number. Therefore, the quadratic function 4x^2 - x + 5 has no real roots.