AS Level Maths Help Please!?

Question

Can someone please help me with these questions? :) my exam board is wjec by the way.

1) The curve C has equation y=x^2+(4k+3)x+7

And the line L has equation y=x+k Where k is a constant

(x being the letter not multiplication sign)

Given that L and C intersect at two distinct points... A) show that 4k^2+5k-6>0 B) find the range of values of k satisfying the inequality

Thanks!! :)

Viola · Accepted Answer

Where they intersect, we can equate the x terms of the equations thus:

x² + (4k + 3)x + 7 = x + k

=> x² + (4k + 2)x + (7 - k) = 0

(A) So for there to be two solutions, the discriminant of this equation must be strictly greater than zero, i.e.

(4k + 2)² - 4(7 - k) > 0

=> 16k² + 16k + 4 - 28 + 4k > 0

=> 16k² + 20k - 24 > 0

=> 4k² + 5k - 6 > 0  QED.

(B) Factorise the above inequality:

(4k  - 3)(k + 2) > 0

So k < 2 or k > 3/4

? · Answer

The line and curve intersect when:

x² + (4k + 3)x + 7 = x + k

so, x² + (4k + 2)x + 7 - k = 0

Using the quadratic formula we get:

x = [-(4k + 2) ± √((4k + 2)² - 4(7 - k))]/2

For there to be two distinct points of intersection the discriminant must be greater than zero.

so, (4k + 2)² - 4(7 - k) > 0

i.e. 16k² + 20k - 24 > 0

=> 4k² + 5k - 6 > 0

so, (4k - 3)(k + 2) > 0

Then, k < -2 or k > 3/4

:)>

Eliot · Answer

x² + (4k + 3)x + 7 = x + k
x² + (4k + 2)x + 7 - k = 0
For two distinct roots the discriminant must be grater than 0
(4k + 2)² - 4(7 - k) > 0
Do the algebra yourself.

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AS Level Maths Help Please!?

3 Answers