x2 + 5x -14 =0
(2, -7)
b2 + 16b + 64
(b +8)2
3y2-5y +2 =0
(2/3, 1)
(x2 - 3) (x + 5)
x3 + 5x2 - 3x - 15
Where is my "a" term located?
In front of x2
x2 + 2x − 1 = 2
(1, -3)
a2 + 11a + 18
(a + 2)(a + 9)
4b2 + 2b - 6 =0
(-3/2, 1)
(4x - 5)(x - 3)
4x2 -17x +15
What is another way to solve some problems that can be solved by the quadratic formula?
Factoring
2x2+9x−5=0
(1/2, -5)
v2 - 7v +10
(v -5)(v -2)
3m2+12m+7=0
m = (-6 +- sqrt 15)/3
3(y+4)(y-7)
3y2-9y-84
In front of what variable would I find my "b" term?
x
k2− 31 − 2k = −6 − 3k2− 2k
(5/2, -5/2)
p2 + 11p + 10
(p + 10)(p + 1)
5b2+2b+4=0
no real solution
(y2 - 2y + 3)(y - 5)
y3 - 7y2 + 13y - 15
What is the quadratic formula used for?
To find the roots, or zeros, of a function
𝑥(𝑥+2)−5=0
X = -1 +- sqrt 6
2n2 + 6n − 108
2(n + 9)(n - 6)
1/9d2 - 1/2d = -1/2
d= 3/2, d= 3
(x - 7)(2x2 + 4)
2x3 - 14x2 +4x - 28
What is the quadratic formula?
-b +- (sqrt b^2-4ac)/2a