Question

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

Answer 1

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

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