# Thread: Stucked @ part ii after comleting part i

1. ## Stucked @ part ii after comleting part i

I don't know how to do the part ii.

The equation of a curve is $y = 4x^2 - 2kx + k$

i) Find the range of values of k if the curve does not meet the x-axis.

ii) Show that the line $y = x + 1$ intersects the curve for all real values of k.

Solution
i) Since curve does not meet x-axis, $b^2-4ac < 0$
$(-2k)^2-4(4)(k) < 0$
$4k^2-16k < 0$
$k(4k-16) < 0$
k < 0 or 4k < 16
k < 4
Range of Values of k is k < 4

ii) Stucked!

2. Does "Real Values" in part ii means that the numbers must be more than or equal to 0?

3. Observe:

We like to find out for wich k the following equation has solutions:
$4x^2-2kx+k = x+1$
$4x^2+(-1-2k)x+(k-1) = 0$

This equation has real solutions if: $D= (-1-2k)^2-16(k-1) = 4k^2-12k+17 \geq 0$

It's not hard to see that D > 0 for all real k.

Does "Real Values" in part ii means that the numbers must be more than or equal to 0?
No it means the numbers are not complex, that is of the form a+bi
where $i = \sqrt{-1}$

4. Originally Posted by Dinkydoe
Observe:

We like to find out for wich k the following equation has solutions:
$4x^2-2kx+k = x+1$
$4x^2+(-1-2k)x+(k-1) = 0$

This equation has real solutions if: $D= (-1-2k)^2-16(k-1) = 4k^2-12k+17 \geq 0$

It's not hard to see that D > 0 for all real k.

No it means the numbers are not complex, that is of the form a+bi
where $i = \sqrt{-1}$
What you have written is not sufficient to say that the Discriminant is always nonnegative.

You have $\Delta = 4k^2 - 12k + 17$

$= 4\left(k^2 - 3k + \frac{17}{4}\right)$

$= 4\left[k^2 - 3k + \left(-\frac{3k}{2}\right)^2 - \left(-\frac{3k}{2}\right)^2 + \frac{17}{4}\right]$

$= 4\left[\left(k - \frac{3k}{2}\right)^2 + \frac{17 + 9k^2}{4}\right]$

$= 4\left(k - \frac{3k}{2}\right)^2 + 17 + 9k^2$.

This is sufficient to say that the discriminant is nonnegative for all $k$.

5. What you have written is not sufficient to say that the Discriminant is always nonnegative
Ofcourse it's not fully sufficient, but a simple argument is:

$\Delta(k)= 4k^2-12k+17 = 0$ has no real solutions
since $\Delta(\Delta(k)) = 144-16\cdot 17 = -128 < 0$

Therefore by continuity of $\Delta(k)$ we have $\Delta(k)> 0$ or $\Delta(k) < 0$ for all k.

Since for all negative k we have $\Delta(k) > 0$ it must be the first case.

6. LOL, I don't understand the question neither do I understand the solutions you guys have provided.. But great job.

From my observation, is the way of completing part ii by subsituting values of k that are < 4?