Which quadratic equation has two irrational roots?
O x2 + x + 3 = 0
O x² + 2x - 8=0
x² + 3x – 1=0
x2 - 2x + 1 = 0

x² + 3x – 1=0

Step-by-step explanation:

hello :

calculate : ∆ = b² - 4ac

x² + x + 3 = 0  :  a =1     b=1     c=3      ∆ = (1)² - 4(1)(3)= -11 < 0 non reals roots

x² + 2x - 8=0  :  a =1     b=2     c=-8     ∆ = (2)² - 4(1)(-8)=36 two rational roots

x² + 3x – 1=0  :  a =1     b=3     c=-1      ∆ = (3)² - 4(1)(-1) = 13 no perfect squard

so : two irrational roots?

x2 - 2x + 1 = 0 : (x-1)²=0      one root

when rotating a figure 180 degrees, imagine that you are able to take the figure and turn it clockwise or counterclockwise around a center point. to rotate a figure 180 degrees, you will need to apply the rule (x, y) → (-x, -y). start by using a coordinate grid with coordinates for each vertex of the figure. the center point of the coordinate grid is located at (0, 0), which is what you will rotate the figure around. write down the original coordinates of the shape you are going to rotate. then, apply the rule. for example, coordinate (1, 2) becomes (-1, -2). another example would be, coordinate (-4, -2) becoming (4, 2). after you change each original coordinate to the rotated coordinate, you will draw your new figure.

step-by-step explanation:

option (b) is correct

step-by-step explanation:

to find the solution of this system, we need to convert it into reduced row-echelon form. we have,

after applying row operations r2 -> r2/2 and r3 -> r3/3, we get,