Answer:
The solution for k is the interval (-3.5,1.5)
Step-by-step explanation:
we have
[tex]k(x^{2}+1)=x^{2}+3x-3[/tex]
[tex]kx^{2}+k=x^{2}+3x-3[/tex]
[tex]x^{2}-kx^{2}+3x-3-k=0[/tex]
[tex]}[1-k]x^{2}+3x-(3+k)=0[/tex]
we know that
If the discriminant is greater than zero . then the quadratic equation has two real and distinct solutions
The discriminant is equal to
[tex]D=b^{2}-4ac[/tex]
In this problem we have
a=(1-k)
b=3
c=-(3+k)
substitute
[tex]D=3^{2}-4(1-k)(-3-k)\\ \\D=9-4(-3-k+3k+k^{2})\\ \\D=9+12+4k-12k-4k^{2}\\ \\D=21-8k-4k^{2}[/tex]
so
[tex]21-8k-4k^{2} > 0[/tex]
solve the quadratic equation by graphing
The solution for k is the interval (-3.5,1.5)
see the attached figure
