Answer:
[tex]-6 < k < 2[/tex]
Step-by-step explanation:
Given
[tex]y = x^2 - 2x[/tex]
[tex]y =kx -4[/tex]
Required
Possible values of k
The general quadratic equation is:
[tex]ax^2 + bx + c = 0[/tex]
Subtract [tex]y = x^2 - 2x[/tex] and [tex]y =kx -4[/tex]
[tex]y - y = x^2 - 2x - kx +4[/tex]
[tex]0 = x^2 - 2x - kx +4[/tex]
Factorize:
[tex]0 = x^2 +x(-2 - k) +4[/tex]
Rewrite as:
[tex]x^2 +x(-2 - k) +4=0[/tex]
Compare the above equation to: [tex]ax^2 + bx + c = 0[/tex]
[tex]a = 1[/tex]
[tex]b= -2-k[/tex]
[tex]c =4[/tex]
For the equation to have two distinct solution, the following must be true:
[tex]b^2 - 4ac > 0[/tex]
So, we have:
[tex](-2-k)^2 -4*1*4>0[/tex]
[tex](-2-k)^2 -16>0[/tex]
Expand
[tex]4 +4k+k^2-16>0[/tex]
Rewrite as:
[tex]k^2 + 4k - 16 + 4 >0[/tex]
[tex]k^2 + 4k - 12 >0[/tex]
Expand
[tex]k^2 + 6k-2k - 12 >0[/tex]
Factorize
[tex]k(k + 6)-2(k + 6) >0[/tex]
Factor out k + 6
[tex](k -2)(k + 6) >0[/tex]
Split:
[tex]k -2 > 0[/tex] or [tex]k + 6> 0[/tex]
So:
[tex]k > 2[/tex] or k [tex]> -6[/tex]
To make the above inequality true, we set:
[tex]k < 2[/tex] or [tex]k >-6[/tex]
So, the set of values of k is:
[tex]-6 < k < 2[/tex]
