Answer:
k ∈ (-∞,[tex]-\frac{1}{2}[/tex])∪(1,∞)
Step-by-step explanation:
For quadratic equations [tex]ax^2+bx+c=0,a\neq 0[/tex] you can find the solutions with the Bhaskara's Formula:
[tex]x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}\\and\\x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}[/tex]
A quadratic equation usually has two solutions.
If you only want real solutions the condition is that the discriminant ([tex]\Delta[/tex]) has to be greater than zero, this means:
[tex]\Delta=b^2-4ac>0[/tex]
Then we have the expression:
[tex](k+1)x^2+4kx+2=0[/tex]
[tex]a=(k+1)\\b=4k\\c=2\\[/tex]
Now to find two distinct real solutions to the original quadratic equation we have to calculate the discriminant:
[tex]b^2-4ac>0\\(4k)^2-4.(k+1).2>0\\16k^2-8(k+1)>0\\16k^2-8k-8>0[/tex]
We got another quadratic function.
[tex]16k^2-8k-8>0[/tex] we can simplify the expression dividing both sides in 8.
[tex]16k^2-8k-8>0\\\\\frac{16k^2}{8} -\frac{8k}{8} -\frac{8}{8} >\frac{0}{8}\\\\2k^2-k-1>0[/tex]
We can apply Bhaskara's Formula except that the condition in this case is that the solutions have to be greater than zero.
[tex]2k^2-k-1>0\\a=2\\b=-1\\c=-1[/tex]
[tex]k_1=\frac{-(-1)+\sqrt{(-1)^2-4.2.(-1)}}{2.2}=\frac{1+\sqrt{9} }{4}=\frac{1+3}{4} =1 \\and\\k_2=\frac{-(-1)-\sqrt{(-1)^2-4.2.(-1)}}{2.2}=\frac{1-3}{4}=-\frac{2}{4}=-\frac{1}{2}[/tex]
Then,
[tex]k>1 \\and\\k<-\frac{1}{2}[/tex]
The answer is:
For all the real values of k who belongs to the interval:
(-∞,[tex]-\frac{1}{2}[/tex])∪(1,∞)
there are two distinct real solutions to the original quadratic equation [tex](k+1)x^2+4kx+2=0[/tex]
