x^2+6x+27
m^2-m-12
What are you supposed to do? Factor?
yes
yes factor
m^2 - m - 12 is simply (x-4)(x+3)
x^2 + 6x + 27 is a weird one if you only just started learning how to factor. Not to mention the factors contain imaginary numbers. You can solve it in two ways. The first way I forgot -.-. The second is to use the quadratic formula which goes (-b+sqrt(b^2-4ac))/(2a) and/or (-b-sqrt(b^2-4ac))/(2a).
I'm going to rewrite the equation as such:
1x^2 + 6x^1 + 27x^0
ax^2 + bx^1 + cx^0
a is the coefficient of x^2 which is 1. b is the coefficient of x which is 6 and c is the coefficient of x^0 which is 27.
So substitute the values of a, b, and c into the quadratic formula and you get the roots. Which are -3 (+/-) 3sqrt(-2). Then subtract them from x and you get your factors which are (x+3-3sqrt(-2)) and (x+3+3sqrt(-2)).
Now that I finished it's ridiculously silly for your teacher to ask this of you if you just learned to factor. Or if you are really behind than you need to study -.- since factoring is basic stuff. Well unless you meant x^2-6x-27 than that's a lot more simpler whose factors are (x-9)(x+3).
To find the solutions for the given quadratic equations, we can use the quadratic formula:
For a quadratic equation of the form ax^2 + bx + c = 0, the solutions can be found using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Let's solve the first equation, x^2 + 6x + 27:
Comparing this equation to the standard form ax^2 + bx + c = 0, we have:
a = 1, b = 6, and c = 27.
Now, substitute these values into the quadratic formula:
x = (-6 ± √(6^2 - 4*1*27)) / (2*1)
Simplifying further:
x = (-6 ± √(36 - 108)) / 2
x = (-6 ± √(-72)) / 2
We notice that the discriminant (√(b^2 - 4ac)) is negative, which means the quadratic equation has no real solutions. The solutions are complex numbers.
For the second equation, m^2 - m - 12:
Comparing this equation to the standard form ax^2 + bx + c = 0, we have:
a = 1, b = -1, and c = -12.
Again, substitute these values into the quadratic formula:
m = (1 ± √((-1)^2 - 4*1*(-12))) / (2*1)
Simplifying further:
m = (1 ± √(1 + 48)) / 2
m = (1 ± √49) / 2
m = (1 ± 7) / 2
So the solutions for the second equation are:
m = (1 + 7) / 2 = 8 / 2 = 4
m = (1 - 7) / 2 = -6 / 2 = -3
Therefore, the solutions for the first equation are complex numbers (no real solutions), and the solutions for the second equation are m = 4 and m = -3.