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LenaWriter [7]
3 years ago
13

Find the set of values of k for which the line y=kx-4 intersects the curve y=x²-2x at 2 distinct points?

Mathematics
1 answer:
Sonbull [250]3 years ago
3 0

Answer:

-6 < k < 2

Step-by-step explanation:

Given

y = x^2 - 2x

y =kx -4

Required

Possible values of k

The general quadratic equation is:

ax^2 + bx + c = 0

Subtract y = x^2 - 2x and y =kx -4

y - y = x^2 - 2x - kx +4

0 = x^2 - 2x - kx +4

Factorize:

0 = x^2 +x(-2 - k) +4

Rewrite as:

x^2 +x(-2 - k) +4=0

Compare the above equation to: ax^2 + bx + c = 0

a = 1

b= -2-k

c =4

For the equation to have two distinct solution, the following must be true:

b^2 - 4ac > 0

So, we have:

(-2-k)^2 -4*1*4>0

(-2-k)^2 -16>0

Expand

4 +4k+k^2-16>0

Rewrite as:

k^2 + 4k - 16 + 4 >0

k^2 + 4k - 12 >0

Expand

k^2 + 6k-2k - 12 >0

Factorize

k(k + 6)-2(k + 6) >0

Factor out k + 6

(k -2)(k + 6) >0

Split:

k -2 > 0 or k + 6> 0

So:

k > 2 or k > -6

To make the above inequality true, we set:

k < 2 or k >-6

So, the set of values of k is:

-6 < k < 2

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Answer:

Step-by-step explanation:

6x - 5y = 15....... (1)

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Putting (2) in (1)

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Step-by-step explanation:

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