Which of the following equations only have one solution? x2+4x+4=0, x2+x=0, x2-1=0

x2+4x+4=0

Step-by-step explanation:

To find the number of real roots for a quadratic, we apply the discriminant. The discriminant is the inside portion of the square root from the quadratic formula.

• yields 2 real roots

• yields 1 real root

•[tx]b^2-4ac<0[/tex] yields no real roots

x2+4x+4=0 has a = 1, b = 4 and c = 4. So 4^2 - 4(1)(4) is 16 - 16 = 0. This yields 1 root.

x2+x=0 has a = 1 and b = 1 and c = 0. So 1^2 - 4(1)(0) = 1- 0 = 1. This yields 2 roots.

x2-1=0 has a = 1 and b = 0 and c= -1. So 0^2 -4(1)(-1) = 0 + 4 = 4. This yields 2 roots.

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2

that is the sum

step-by-step explanation:

there are no other answers without knowing the variables values