x^2-4x+32=0.
would the final answer be 2 +or- 2 sqrt7 ?
x^2-4x+32=0
a =1
b= -4
c= 32
4 +or- sq root (-4)^2 -4(1)(32)
4 +or- sq root 16-128
4 +or- sq root -112
you can't take the sq root of a (-) number so you can't solve this :/
How did you get 2 +or- 2 sqrt7.
Perhaps I went wrong somewhere.
hmmm. what's b^2-4ac? 16-128 = -112
So, x = 2 ± 2√(-7) = 2 ± 2√7 i
gotta watch those pesky minus signs
the original question was
x^2-4x=32
can you help me solve?
as I suspected, that makes it x^2-4x-32=0
since 32=8*4,
(x-8)(x+4) is the factorization
To find the solutions of the quadratic equation x^2 - 4x + 32 = 0, you can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, a = 1, b = -4, and c = 32.
Substituting these values into the quadratic formula, we have:
x = (-(-4) ± √((-4)^2 - 4(1)(32))) / (2(1))
= (4 ± √(16 - 128)) / 2
= (4 ± √(-112)) / 2
Since the value inside the square root (√) is negative, it means that the quadratic equation has no real solutions. The square root of a negative number does not yield a real number.
Therefore, the final answer for x in this case would be that there are no real solutions, or in other words, the quadratic equation does not intersect the x-axis.