Anyone can help.. I'm stuck with this question..?

Question

The line (k-2)y = 3x meets the curve xy = 1-x at two distinct points. Find the range of k.

Answer 1

k<-10 or k>2

(k-2)y = 3x or
y = 3x/(k-2)
and
xy = 1-x or
y = (1-x)/x

so
3x/(k-2) = (1-x)/x
3x^2 = (1-x)(k-2)
3x^2 + (k-2)x - (k-2) = 0
3x^2 + (k-2)x + (2-k) = 0
This will have two real roots when b^2-4ac>0 or:

(k-2)^2 - 4(3)(2-k) > 0
k^2 - 4k + 4 - 24 + 12k > 0
k^2 + 8k - 20 > 0
(k+10)(k-2) > 0
The left side will be positive when both terms are positive or both are negative. They are both positive when k>2, and they are both negative when k<-10.

So the final solution is k>2 or k<-10

For those of you who said k<-2, try k = -4.
Then the line is y = -x/2. This line does not intersect the curve at all.

Answer 2

The line (k-2)y = 3x meets the curve xy = 1-x at two distinct points. Find the range of k.

xy = 1 - x
So (k - 2)xy = 3(k - 2)(1 - x)

so x * 3x = 3(k - 2) - 3(k - 2)x

Whence x² + (k - 2)x - (k - 2) = 0

Now this quadratic equation has two distinct real roots if Δ > 0 ie if b² - 4ac > 0)

ie if (k - 2)² - 4*-(k - 2) > 0
ie k² - 4k + 4 + 4k - 8 >0
ie if k² - 4 > 0
ie k < -2 or k > 2

Answer 3

(k-2)y=3x.........(i)
xy=1-x.............(ii)

At the point of intersection the value of x, y will be same

=>y=3x/(k-2)...........from (i)

Substitute in (ii)

=>x(3x/(k-2))=1-x

=>3x^2=(1-x)(k-2)
=>3x^2=-kx + 2x + k - 2
=>3x^2 + x(k-2) + 2-k = 0

For real values of x, the discriminant of the equation should be >=0, also given that values of x are distinct therefore roots will never be equal <=> discriminant >0

=>(k-2)^2 - 4(3 * (2-k))>0
=>k^2 + 4 -4k -4(6 - 3k)>0
=>k^2 + 4 -4k -24 + 12k>0
=>k^2 + 8k - 20>0
=>k^2 + 10k - 2k -20>0
=>k(k+10) - 2(k+10)>0
=>(k+10)(k-2)>0

=>k ε(-∞,-10)U(2,∞).....................ε means belongs to

Answer 4

k>2 or k<-10

Answer 5

xy = 1 - x
So (k - 2)xy = 3(k - 2)(1 - x)

so x * 3x = 3(k - 2) - 3(k - 2)x

Whence x² + (k - 2)x - (k - 2) = 0

Now this quadratic equation has two distinct real roots if Δ > 0 ie if b² - 4ac > 0)

ie if (k - 2)² - 4*-(k - 2) > 0
ie k² - 4k + 4 + 4k - 8 >0
ie if k² - 4 > 0
ie k < -2 or k > 2