Solving Discriminant Question & Finding Possible Values of k

[_Mayday_]

[SOLVED] Discriminant Question

The Equation:

$x^2 + kx + 8 = k$

has no real roots

(a) Show that $k$ satisfies $k^2 + 4k - 32 (is.smaller.than) 0$

(b) Hence find the set of possible values of $k$Attempt

(a) First I will put it in the form:
$ax^2 + bx + c = 0$
$x^2 + kx + 8 = k$

a = 1, b = k, c = 8

Now I will put this into the discriminant.

$b^2 - 2ac (is.smaller.than) k$
$k^2 - 16 (is.smaller.than) k$

Where do I go from here??

Thanks for any help!_Mayday_

 

[Hootenanny]

(a) First I will put it in the form:
$ax^2 + bx + c = 0$
$x^2 + kx + 8 = k$

a = 1, b = k, c = 8

You have said that you were going to put it into canonical form, but haven't done it! Your RHS should be zero.

 

[_Mayday_]

What happens to the k then?

 

[Hootenanny]

What happens to the k then?

$$x^2+kx+8-k=0$$

 

[_Mayday_]

:rofl:

$$x^2 + kx + 8 - k$$

Ok, but now how do I know which bit is which?

a = 1, b = k, c = 8-k?

 

[Hootenanny]

:rofl:

$$x^2 + kx + 8 - k$$

Ok, but now how do I know which bit is which?

a = 1, b = k, c = 8-k?

Sounds good to me

 

[_Mayday_]

$$b^2 - 2ac$$
$$k^2 - 2(8 - k)$$
$$k^2 - 16 + 2k = 0$$
$$k^2 - 2k = - 16$$

Where do I go from here? I need to prove that the LHS is smaller than 0.

 

[Hootenanny]

$$b^2 - {\color{red}2}ac$$

:uhh:

 

[_Mayday_]

Dammit, how embaressing.

$$b^2 - 4ac$$
$$k^2 - 4(8 - k)$$
$$k^2 - 32 + 4k = 0$$
$$k^2 + 4k = 32$$

 

[Hootenanny]

Dammit, how embaressing.

$$b^2 - 4ac$$
$$k^2 - 4(8 - k)$$
$$k^2 - 32 + 4k = 0$$

This line is a good place to stop. Now consider the quadratic equation, what is the condition for there to be real roots?

 

[_Mayday_]

The question says there are no real roots. For there to be real roots the discriminant need to be equal to or larger than 0.

 

[Hootenanny]

For there to be real roots the discriminant need to be equal to or larger than 0.

Correct, so for non-real roots...

 

[_Mayday_]

The discriminant must be smaller than 0.

 

[Hootenanny]

The discriminant must be smaller than 0.

Correct.

 

[_Mayday_]

So then my answer is:

$$k^2 - 32 + 4k (smaller.than) 0$$

??

 

[Hootenanny]

So then my answer is:

$$k^2 - 32 + 4k (smaller.than) 0$$

??

Indeed it is, which is what you were asked to prove.

P.S. Latex allows use of '<' symbols.

 

[_Mayday_]

Haha what a relief! Never knew > worked!

Now to find possible values I simply factorise?

$$(k - 4)(k + 8)$$

So the possible values are 4 and -8?

 

[Hootenanny]

Haha what a relief! Never knew > worked!

Now to find possible values I simply factorise?

$$(k - 4)(k + 8)$$

So the possible values are 4 and -8?

No these values provide you with the range of an interval. Now you need to decide whether the segment you want (i.e. the set of numbers which yield complex roots) lies within this interval or outside it.

 

[_Mayday_]

Hoot, I got to dash. I'll be back on in an hour or so, see you if you're on. Thanks for your help upto now though!

 

[_Mayday_]

No these values provide you with the range of an interval. Now you need to decide whether the segment you want (i.e. the set of numbers which yield complex roots) lies within this interval or outside it.

Sorry Hoot, I am struggling to understand what that means

 

[HallsofIvy]

(k-4)(k+8) is equal to 0 when k= -8 or 4. If the product of two numbers is negative, what is must be true of the numbers? (+ times+ = + , - time -= + so ?)

 

[_Mayday_]

If the product is negative, then one needs to be negative.

 

[HallsofIvy]

And the other must be positive. Any number that satisfies k- 4> 0 also satisfies k+ 8> 0. Any number that satisfies k+ 8< 0 also satisfies k-4< 0. What numbers satisy k- 4< 0, k+ 8> 0?

 

[_Mayday_]

-8 < k < 4

 

[Hootenanny]

-8 < k < 4

Sounds good to me

 

[_Mayday_]

Cheers Hoot and Halls!

 

[HallsofIvy]

Yes. Now what value of k make that smaller than 0?


Hint: value of k that make it equal to 0 separate those that make it larger than 0 and those that make it less than 0.