p2 + -2p - 10 = 5
5, -3
4x2 + 25 = 125
x= 5 or x = -5
m2 − 5m − 14 = 0
{7, −2}
a2 + 14a - 51 = 0
{3, -17}
What value of c would make this a perfect square trinomial?
x^2+4x+c
16
9n2 + 39n = -36
-4/3, -3
(4x + 1)2 - 16 = 0
x = 3/4 or x = -5/4
How many and what type of solutions comes from this equation?
4n^2-4n-24
2 real
x2 − 12x + 11 = 0
{11, 1}
What value of c would make this a perfect square trinomial?
s^2-26x+c
c=169
7r2 + 84 = -49r
-4, -3
34 = (a - 2)2 +66
a=+-4isqrt2 +2
2x2 − 3x − 5 = 0
{5/2 , −1}
n2 = 18n + 40
{20, −2}
x^2+4x-2
-2+-sqrt6
3v2 + 7v = 40
8/3, -5
0 = 3(x + 7)2 - 24
-7+-2sqrt2
9n2 = 4 + 7n
(7+-sqrt193)/18
x2 − 10x + 26 = 8
{5 + square root 7, 5 square root − 7}
5+-sqrt7
x^2+6x+3
-3+-sqrt6
a^2-49=0
a=+-7
Simplify (3-2i) - 2i(8-6i)
-9-18i
Find a possible pair of integer values for a and c so that the equation has 2 imaginary solutions.
ax^2+4x+c=0
1,5 (sample)
7(x-4)^2-18=10
x=6,2
Convert to vertex form and find the vertex.
s^2+2s-6
(-2,-10)