what values of K will roots of 2x^2+Kx+1=0 be real?

For the roots of a quadratic equation to be real, its discriminant b^2 - 4ac has to be greater than or equal to 0 (it can't be negative). In this equation, a = 2, b = k, and c = 1. Solve for k.

b^2 - 4ac ≥ 0

k^2 - 4(2)(1) ≥ 0

k^2 - 8 ≥ 0

k^2 ≥ 8

k ≤ -√8 or k ≥ √8

k ≤ -2√2 or k ≥ 2√2 <===ANSWER

2x^2+Kx+1=0

x^2 + kx/2 = -1/2 ### divide each term by 2

x^2 + kx/2 + (k/4)^2 = (k/4)^2 - 1/2 #### complete the square

(x + k/4)^2 = (k/4)^2 - 1/2 = k^2/16 -1/2 = k^2/16 - 8/16 = (k^2-8 ) /16

(x + k/4)^2 =(k^2-8 ) /16

Take square roots of both sides

x + k/4 = +/- (k^2-8)^1/2 / 4

x = - k/4 +/- (k^2-8)^1/2 / 4

The roots are real when k^2 - 8 >=0

k^2 >= 8

k <= - 8^1/2 or k >= 8^1/2

Use the quadratic formula:

x = -k +/- sqrt(k^2 - 4*2*1)

The roots will be real when the part inside the square root are >0, so

k^2 > 8 or k > 2*sqrt(2) (jumped a few steps in the algebra, you fill them in)

for actual roots, you want the area of the quadratic formula below the sq. root to be positive that component is b^2 - 4ac. plugging on your values you get ok^2 - 4*2*a million this value needs to be more suitable than 0 or equivalent to it so ok^2 - 8 >= 0 upload 8 to each and each and every part ok^2 >= 8 take root ok >= root(8) or ok<= -root(8)