Question 1100749
<pre><b>
{{{2x^2-kx+2}}}{{{""=""}}}{{{0}}}

The discriminant, which is b<sup>2</sup>-4ac, must be zero in 
order for this equation to have just one real root:

{{{discriminant}}}{{{""=""}}}{{{b^2-4ac}}}{{{""=""}}}{{{(-k)^2-4(2)2)}}}{{{""=""}}}{{{k^2-16}}}

So we set this discriminant equal to zero:

{{{k^2-16}}}{{{""=""}}}{{{0}}}

Solve that either of two ways:

First way:

{{{k^2-16}}}{{{""=""}}}{{{0}}}

Add 16 to both sides:

{{{k^2}}}{{{""=""}}}{{{16}}}

Use the principle of square roots:

{{{k}}}{{{""=""}}}{{{"" +- 4}}}   <-- the answers 

Second way:

{{{k^2-16}}}{{{""=""}}}{{{0}}}

Factor the left side as the difference of two squares:

{{{(k-4)(k+4)}}}{{{""=""}}}{{{0}}}

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

-------

To prove that {{{2x^2-kx+2}}}{{{""=""}}}{{{0}}} just has one
real root when k = 4 or k = -4, we substitute in

{{{2x^2-kx+2}}}{{{""=""}}}{{{0}}}

Substituting 4 for k:

{{{2x^2-4x+2}}}{{{""=""}}}{{{0}}}

Divide both sides by 2

{{{x^2-2x+1}}}{{{""=""}}}{{{0}}}

Factor:

{{{(x-1)(x-1)}}}{{{""=""}}}{{{0}}}

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

---

{{{2x^2-kx+2}}}{{{""=""}}}{{{0}}}

Substituting -4 for k:

{{{2x^2-(-4)x+2}}}{{{""=""}}}{{{0}}}

{{{2x^2+4x+2}}}{{{""=""}}}{{{0}}}

Divide both sides by 2

{{{x^2+2x+1}}}{{{""=""}}}{{{0}}}

Factor:

{{{(x+1)(x+1)}}}{{{""=""}}}{{{0}}}

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

Edwin</pre>