Question

For which k are the roots of k(x2+1)=x2+3x–3 real and distinct?

Answer 1

Answer:

The solution for k is the interval (-3.5,1.5)

Step-by-step explanation:

we have

$k(x^{2}+1)=x^{2}+3x-3$

$kx^{2}+k=x^{2}+3x-3$

$x^{2}-kx^{2}+3x-3-k=0$

$}[1-k]x^{2}+3x-(3+k)=0$

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

$D=b^{2}-4ac$

In this problem we have

a=(1-k)

b=3

c=-(3+k)

substitute

$D=3^{2}-4(1-k)(-3-k)\\ \\D=9-4(-3-k+3k+k^{2})\\ \\D=9+12+4k-12k-4k^{2}\\ \\D=21-8k-4k^{2}$

so

$21-8k-4k^{2} > 0$

solve the quadratic equation by graphing

The solution for k is the interval (-3.5,1.5)

see the attached figure