find the value of ''k'' given that one solution of the equation

Question

5x²-kx+40=0

 

Answer 1

For the solution to be one (actually two equal reals) the discriminant must be equal to zero.

b² - 4(a)(c) = 0

k² - 4(5)(40) = 0

k² - 800 = 0

k² = 800

k = ±20√2

Answer 2

5x²- kx + 40 = 0

 

find the value of ''k'' given that there is one solution to the equation

 

5(x² - kx/5 + 8) = 0

 

x² - kx/5 + 8 = 0

 

If there is one solution for x then the quadratic equation, ax² +bx + c = 0 must be in the form (x ± √c)² = 0

 

(x - √8)² = 0

 

(x - 2√2)(x - 2√2) = x² - 2√2(x) - 2√2(x) + 8

 

x² - 4√2(x) + 8

 

Then

 

-k/5 = -4√2

 

k = 20√2

 

k can actually be ± 20√2 for there to only be one solution for x