# Given that L intersects C in two distinct points, show that k>2?

Help with this question would be very much appreciated. I can not find a question like this in my books.The question is as follows;

The curve C has equation y = 4x^2 - 7x + 11

and the line L has equation y = 5x + k

where k is a constant.

Given that L intersects C in two distinct points, show that k > 2.

Relevance
• To have two points of intersection on the (x,y) coordinate plane, the two equations must be solved simultaneously to find pairs of numbers (x,y) for which both the curve's equation and the line's equation are true.

So assume x is a number, y is a number, y = 4x^2 - 7x + 11, and y = 5x+k are true.

Now algebraic operations can be done to both equations together.

Geometrically what is happening is that C is a parabola opening upwards, with a minimum. If k is too small then the line will pass underneath the parabola. If k is just right, the line will be tangent to the parabola and touch it in one point. If k is a little bigger, then the line will intersect the parabola in two points.

Solve y = 4x^2 - 7x + 11 = 5x + k

There will be a quadratic equation and the solution will have two values only when the discriminant is positive. Or completing the square, the square will be on the left side and the solution will have two values only when the right side is positive.

4x^2 - 12x + 11 - k = 0

Divide by 3

x^2 - 4x + (11-k)/3 = 0

Complete the square by adding 4 to both sides

(x-2)^2 = 12/3 - 11/3 + k/3 = (1+k)/3

1+k > 3

k > 2

• Equation of Curve C : y = 4x² - 7x + 11

Equation of Line L : y = 5x + k

Line L intersects the Curve C in two distinct points

=> they have two common points of intersection

=> Coordinates of the common points will satisfy both the equations

(p, q) be the coordinates of the common points

Hence, q = 4p² - 7p + 11

and q = 5p + k

=> 4p² - 7p + 11 = 5p + k

=> 4p² - 12p = k - 11

=> (2p)² - 2(2p)(3) + 3² - 3² = k - 11

=> { (2p - 3)²} - 9 = k - 11

=> (2p - 3)² = k - 11 + 9 = k - 2

=> (2p - 3) = ± √(k - 2)

In order to have 2 distinct values of p and also of q, the term under the sqrt sign

on the RHS must be positive

=> k - 2 > 0

=> k > 2

• we can find the point of intersection by subsituting the second equation into the first.

5x + k = 4x^2 -7x +11.

0= 4x^2 - 12x +11 - k

Now discriminant > 0 ( as there are 2 roots).

144 - 4*4*(11- k)>0

9-(11-k)>0

K>2.

• y = 4x^2 - 7x + 11

y = 5x + k

to find where they intersect solve them simultaneously - equate the y's

5x + k = 4x^2 - 7 x + 11

4x^2 - 12 x + 11-k = 0

this quadratic will have two roots if the discriminate is > 0

a = 4: b = 12: c = 11 - k

b^2 - 4 ac = 144 -4(4)(11-k) > 0

144 - 176 + 16k >0

16 k > 32

k > 2

• 4x^2-7x+11 = 5x +k

4x^2 -12x +11-k = 0

D = 12^2 - (4)(4)(11-k) <---- Find Discriminant

D = 144 -176 + 16k = -32 + 16k

To have two intersections D > 0

Thus 16k > 2

k > 2

• To get aspect (or factors) of intersection between curve and line, equate both equations to get: 4x^2 - 7x + 11 = 5x + ok Now rearranging above to get quadratic equation : 4x^2 - 12x + 11- ok =0 Now we choose 2 different factors of intersection, i.e., both roots of above equation are actual and different. for this reason the determinant of the above quadratic equation D > 0 . (be conscious that D=0 for unmarried root which may were the case of in easy words one aspect of intersection) Now D = (-12)^2 -4*4*(11-ok) > 0 which factors ok > 0. be conscious that Determinant D = b^2 -4ac for quadratic equation ax^2 +bx+c=0 i desire it enables

• y = 4x^2 - 7x + 11

y = 5x + k

-----> 4x^2 - 7x + 11 = 5x + k

4x^2 - 7x + 11 - 5x - k = 0

4x^2 - 12x + (11-k) = 0

To get two different x values, we need discriminant > 0

b^2 - 4ac > 0

144 - 16(11-k) > 0

144 - 176 + 16k > 0

-32 + 16k > 0

16k > 32

k > 2

Mαthmφm