SOLUTION: Find all roots of "k" such that 2x^2-kx+2=0 has one real root

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Question 1100749: Find all roots of "k" such that 2x^2-kx+2=0 has one real root
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!

2x%5E2-kx%2B2%22%22=%22%220

The discriminant, which is b2-4ac, must be zero in 
order for this equation to have just one real root:

discriminant%22%22=%22%22b%5E2-4ac%22%22=%22%22%28-k%29%5E2-4%282%292%29%22%22=%22%22k%5E2-16

So we set this discriminant equal to zero:

k%5E2-16%22%22=%22%220

Solve that either of two ways:

First way:

k%5E2-16%22%22=%22%220

Add 16 to both sides:

k%5E2%22%22=%22%2216

Use the principle of square roots:

k%22%22=%22%22%22%22+%2B-+4   <-- the answers 

Second way:

k%5E2-16%22%22=%22%220

Factor the left side as the difference of two squares:

%28k-4%29%28k%2B4%29%22%22=%22%220

k-4 = 0;  k+4 = 0
  k = 4;    k = -4   <-- the answers

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To prove that 2x%5E2-kx%2B2%22%22=%22%220 just has one
real root when k = 4 or k = -4, we substitute in

2x%5E2-kx%2B2%22%22=%22%220

Substituting 4 for k:

2x%5E2-4x%2B2%22%22=%22%220

Divide both sides by 2

x%5E2-2x%2B1%22%22=%22%220

Factor:

%28x-1%29%28x-1%29%22%22=%22%220

x-1 = 0; x-1 = 0
  x = 1   x = 1   <-- just 1 root, 1

---

2x%5E2-kx%2B2%22%22=%22%220

Substituting -4 for k:

2x%5E2-%28-4%29x%2B2%22%22=%22%220

2x%5E2%2B4x%2B2%22%22=%22%220

Divide both sides by 2

x%5E2%2B2x%2B1%22%22=%22%220

Factor:

%28x%2B1%29%28x%2B1%29%22%22=%22%220

x+1 = 0;  x+1 = 0
  x = -1    x = -1   <-- just 1 root, -1

Edwin