Respuesta :
Answer:
If [tex]k = 1[/tex] or [tex]k = -\frac{1}{2}[/tex], there is only one solution to the given quadratic equation.
Step-by-step explanation:
Given a second order polynomial expressed by the following equation:
[tex]ax^{2} + bx + c, a\neq0[/tex]
This polynomial has roots [tex]x_{1}, x_{2}[/tex] such that [tex]ax^{2} + bx + c = (x - x_{1})*(x - x_{2})[/tex], given by the following formulas:
[tex]x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}[/tex]
[tex]x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}[/tex]
[tex]\bigtriangleup = b^{2} - 4ac[/tex]
The signal of [tex]\bigtriangleup[/tex] determines how many real roots an equation has:
[tex]\bigtriangleup > 0[/tex]: Two real and different solutions
[tex]\bigtriangleup = 0[/tex]: One real solution
[tex]\bigtriangleup < 0[/tex]: No real solutions
In this problem, we have the following second order polynomial:
[tex](k+1)x^{2} + 4kx + 2 = 0[/tex].
This means that [tex]a = k+1; b = 4k; c = 2[/tex]
It has one solution if
[tex]\bigtriangleup = 0[/tex]
[tex]b^{2} - 4ac = 0[/tex]
[tex]16k^{2} -8(k+1) = 0[/tex]
[tex]16k^{2} - 8k - 8 = 0[/tex]
We can simplify by 8
[tex]2k^{2} - k - 1 = 0[/tex]
The solution is:
[tex]k = 1[/tex] or [tex]k = -\frac{1}{2}[/tex]
So, if [tex]k = 1[/tex] or [tex]k = -\frac{1}{2}[/tex], there is only one solution to the given quadratic equation.
