which quadratic formula would be used to solve the following quadratic equation?
4x^2 - 5x = -2
can you show me the correct answer and how do it please
I would use the standard quadratic formula for ax²+bx+c=0, and
x=(-b±√(b²-4ac))/2a
Here
4x^2 - 5x = -2
can be rewritten as
4x^2 - 5x + 2 =0, so
a=4,
b=-5
c=2
substituting in the quadratic formula,
x=(-(-5)±√((-5)²-4(4)(2)))/(2*4)
which when simplified, gives
x=(5±(√7)i)/8 where i=√(-1) (complex root).
Complex numbers is a little beyond high school, but the procedure is the same.
If you are not expecting complex roots, you may want to recheck if the question you posted was correct. If not, repeat the exercise using the correct question.
using the quadratic formula, how do you solve 2p^2 + 16p = -2p^2 + 3?
To solve the quadratic equation 4x^2 - 5x = -2, we need to use the quadratic formula. The quadratic formula is used to find the values of x that satisfy any quadratic equation in the form ax^2 + bx + c = 0, where a, b, and c are constants.
The quadratic formula is given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, the equation is 4x^2 - 5x = -2. We can rearrange it to get it in the standard form ax^2 + bx + c = 0:
4x^2 - 5x + 2 = 0
From the equation, we can identify the values of a, b, and c:
a = 4
b = -5
c = 2
Now, we can substitute these values into the quadratic formula and solve for x:
x = (-(-5) ± √((-5)^2 - 4*4*2)) / (2*4)
Simplifying further:
x = (5 ± √(25 - 32)) / 8
x = (5 ± √(-7)) / 8
Since the expression inside the square root is negative, the quadratic equation has no real solutions. The ± symbol indicates that there are two imaginary solutions.