# Discriminant of Roots - Stucked @ part ii after completing part i

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Dec 30th 2009, 06:48 PM
Punch
Discriminant of Roots - Stucked @ part ii after completing part i
I don't know how to do the part ii.

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

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

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

Solution
i) Since curve does not meet x-axis, $\displaystyle b^2-4ac < 0$

$\displaystyle -2k^2-4(4)(k) < 0$

$\displaystyle -4k^2-16k < 0$

$\displaystyle k(-4k-16) < 0$

$\displaystyle k < 0$ or $\displaystyle -4k < 16 k < -4$

Range of Values of $\displaystyle k$ is $\displaystyle k < -4$

ii) Stucked!
• Dec 30th 2009, 08:08 PM
sym0110
Quote:

Originally Posted by Punch
I don't know how to do the part ii.

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

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

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

Solution
i) Since curve does not meet x-axis, $\displaystyle b^2-4ac < 0$
$\displaystyle -2k^2-4(4)(k) < 0$$\displaystyle -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!

Suppose the curve and the line do intersect at some point, that means there x and y coordinates at that point are the same, hence we may substitute y=x+1 to the quadratic:
$\displaystyle x+1=4x^2-2kx+k$, rearranging gives:
$\displaystyle 4x^2-(2k+1)x+(k-1)=0$, now consider the discriminant of this quadratic:
$\displaystyle delta=(2k+1)^2-16(k-1)$
$\displaystyle delta=4((k-3/2)^2-7/4)$
implies that when k=3/2, delta<0, hence contradicts our assumption.
• Dec 31st 2009, 12:01 AM
Bacterius
Hello sym0110,
for delta, you can use the \Delta command : $\displaystyle \Delta$
:)

For the second question, you must show that for any $\displaystyle k$, the equation $\displaystyle x + 1 = 4x^2 - 2kx + k$ has at least one real solution in $\displaystyle x$. Check the discriminant of this quadratic equation to see if it can actually have real solutions (negative discriminant means no real solution). Can you follow up ?

Take your quadratic equation : $\displaystyle 4x^2 - 2kx + k - x - 1 = 0$.

This can be factorized as : $\displaystyle 4x^2 - (2k + 1)x + (k - 1) = 0$.

Take the discriminant : $\displaystyle \Delta = b^2 - 4ac = (2k + 1)^2 - 4 \times 4 \times (k - 1)$.

We then have : $\displaystyle \Delta = (2k + 1)^2 - 16(k - 1)$.

We want to prove that $\displaystyle \Delta > 0$, so $\displaystyle (2k + 1)^2 - 16(k - 1) > 0$.

That is : $\displaystyle (2k)^2 + 4k + 1^2 - 16k + 16 > 0$

Simplify further : $\displaystyle 4k^2 + 4k + 1 - 16k + 16 > 0$

Keep going : $\displaystyle 4k^2 - 12k + 17 > 0$

Is this correct ? Let us check the discriminant of this new quadratic :

$\displaystyle \Delta ' = 12^2 - 4 \times 4 \times 17 = 144 - 272 = -128 \ (< 0)$.

If the discriminant is negative, it means the equation has no real roots and thus all values are greater than zero (this is because the $\displaystyle a$ value of the quadratic equation is positive. If it was negative, then all values would be less than zero. Graph some equations to see what I mean). Thus we have proved that :

$\displaystyle 4k^2 - 12k + 17 > 0$

And therefore we have proved that for any real $\displaystyle k$, the line intersects the curve (finish the conclusion properly).
• Dec 31st 2009, 06:23 PM
Punch
Quote:

Originally Posted by Bacterius
Hello sym0110,
for delta, you can use the \Delta command : $\displaystyle \Delta$
:)

For the second question, you must show that for any $\displaystyle k$, the equation $\displaystyle x + 1 = 4x^2 - 2kx + k$ has at least one real solution in $\displaystyle x$. Check the discriminant of this quadratic equation to see if it can actually have real solutions (negative discriminant means no real solution). Can you follow up ?

Take your quadratic equation : $\displaystyle 4x^2 - 2kx + k - x - 1 = 0$.

This can be factorized as : $\displaystyle 4x^2 - (2k + 1)x + (k - 1) = 0$.

Take the discriminant : $\displaystyle \Delta = b^2 - 4ac = (2k + 1)^2 - 4 \times 4 \times (k - 1)$.

We then have : $\displaystyle \Delta = (2k + 1)^2 - 16(k - 1)$.

We want to prove that $\displaystyle \Delta > 0$, so $\displaystyle (2k + 1)^2 - 16(k - 1) > 0$.

That is : $\displaystyle (2k)^2 + 4k + 1^2 - 16k + 16 > 0$

Simplify further : $\displaystyle 4k^2 + 4k + 1 - 16k + 16 > 0$

Keep going : $\displaystyle 4k^2 - 12k + 17 > 0$

Is this correct ? Let us check the discriminant of this new quadratic :

$\displaystyle \Delta ' = 12^2 - 4 \times 4 \times 17 = 144 - 272 = -128 \ (< 0)$.

If the discriminant is negative, it means the equation has no real roots and thus all values are greater than zero (this is because the $\displaystyle a$ value of the quadratic equation is positive. If it was negative, then all values would be less than zero. Graph some equations to see what I mean). Thus we have proved that :

$\displaystyle 4k^2 - 12k + 17 > 0$

And therefore we have proved that for any real $\displaystyle k$, the line intersects the curve (finish the conclusion properly).

Sorry but probably I suck at maths, I have highlighted the few points which I do not understand in red.
• Dec 31st 2009, 06:26 PM
Punch
I do not understand why we have to prove $\displaystyle b^2-4ac>0$.
And what the checking of discriminant is all about, neither do I know where $\displaystyle \Delta ' = 12^2 - 4 \times 4 \times 17 = 144 - 272 = -128 \ (< 0)$ came from.
• Dec 31st 2009, 06:28 PM
Bacterius
Then you should read this prior to messing with quadratic equations :

Quadratic equation - Wikipedia, the free encyclopedia
• Jan 1st 2010, 03:44 AM
Punch
Quote:

Originally Posted by Bacterius
Then you should read this prior to messing with quadratic equations :

Quadratic equation - Wikipedia, the free encyclopedia

Okay, firstly, the part of bring x+1 into the equation was understandable because they have to meet, thus we have to equate them.

However, does x+1=0 means thats it has real roots? I just don't get this part... sorry.
• Jan 7th 2010, 04:49 PM
Boyrog
@Integral
When k=3 the line also intersects the curve, the proof is down there.

@Punch the first part you didnt understand:
This determinant came from the first equation assuming its intersections to the line y=x+1. You need to prove the determinant is bigger than 0 for you dont get a sqrt of negative number, what makes impossible the equation in reals and also impossible the intersection, since that is what we want to proof.

Quote:

Keep going : http://www.mathhelpforum.com/math-he...9c612744-1.gif

Is this correct ? Let us check the discriminant of this new quadratic :

http://www.mathhelpforum.com/math-he...26a76365-1.gif.
The development of our determinant gave another quadratic function and we need to prove now that this function return ONLY positive values. That is, it must be above the x-axis. What quadratics have only positives results? Those with a>0 and roots with complex roots, that is, determinant of the equation must be <0.
This is what Bacterius showed. The equation has only positive values.
I attached the graph of 4kČ-12k+17, and there you can see it returns only positive values.
• Jan 8th 2010, 03:21 AM
HallsofIvy
Quote:

Originally Posted by Punch
I don't know how to do the part ii.

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

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

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

Solution
i) Since curve does not meet x-axis, $\displaystyle b^2-4ac < 0$

$\displaystyle -2k^2-4(4)(k) < 0$

Your "b" is -2k so $\displaystyle b^2= (-2k)^2= 4k^2$ NOT "$\displaystyle -2k^2$".

Quote:

$\displaystyle -4k^2-16k < 0$

$\displaystyle k(-4k-16) < 0$

$\displaystyle k < 0$ or $\displaystyle -4k < 16 k < -4$

Range of Values of $\displaystyle k$ is $\displaystyle k < -4$

ii) Stucked!
• Jan 14th 2010, 01:15 AM
Punch
Quote:

Originally Posted by Bacterius
Hello sym0110,
for delta, you can use the \Delta command : $\displaystyle \Delta$
:)

For the second question, you must show that for any $\displaystyle k$, the equation $\displaystyle x + 1 = 4x^2 - 2kx + k$ has at least one real solution in $\displaystyle x$. Check the discriminant of this quadratic equation to see if it can actually have real solutions (negative discriminant means no real solution). Can you follow up ?

Take your quadratic equation : $\displaystyle 4x^2 - 2kx + k - x - 1 = 0$.

This can be factorized as : $\displaystyle 4x^2 - (2k + 1)x + (k - 1) = 0$.

Take the discriminant : $\displaystyle \Delta = b^2 - 4ac = (2k + 1)^2 - 4 \times 4 \times (k - 1)$.

We then have : $\displaystyle \Delta = (2k + 1)^2 - 16(k - 1)$.

We want to prove that $\displaystyle \Delta > 0$, so $\displaystyle (2k + 1)^2 - 16(k - 1) > 0$.

That is : $\displaystyle (2k)^2 + 4k + 1^2 - 16k + 16 > 0$

Simplify further : $\displaystyle 4k^2 + 4k + 1 - 16k + 16 > 0$

Keep going : $\displaystyle 4k^2 - 12k + 17 > 0$

Is this correct ? Let us check the discriminant of this new quadratic :

$\displaystyle \Delta ' = 12^2 - 4 \times 4 \times 17 = 144 - 272 = -128 \ (< 0)$.

If the discriminant is negative, it means the equation has no real roots and thus all values are greater than zero (this is because the $\displaystyle a$ value of the quadratic equation is positive. If it was negative, then all values would be less than zero. Graph some equations to see what I mean). Thus we have proved that :

$\displaystyle 4k^2 - 12k + 17 > 0$

And therefore we have proved that for any real $\displaystyle k$, the line intersects the curve (finish the conclusion properly).

Hi, is there a specific way in which i must express the equation as?
Do i have to express it as $\displaystyle ax^2-bx+c$ or $\displaystyle ax^2+bx+c$ or is it fine for both?
• Jan 14th 2010, 02:13 AM
Prove It
Quote:

Originally Posted by Punch
Hi, is there a specific way in which i must express the equation as?
Do i have to express it as $\displaystyle ax^2-bx+c$ or $\displaystyle ax^2+bx+c$ or is it fine for both?

Express every Quadratic equation as $\displaystyle ax^2 + bx + c$.
• Jan 14th 2010, 03:44 AM
mr fantastic
Quote:

Originally Posted by Prove It
Express every Quadratic equation as $\displaystyle ax^2 + bx + c$.

Unfortunately, this advice needs to be taken with a grain of salt. However, I'm not going to open up the can of worms by saying that sometimes a factorised form or a turning point form might be more desirable than the standard form.

@OP: You have said in a number of threads that you are bad at maths etc. That's no excuse for not investing the necessary time and effort so that you have a basic understanding of things like the discriminant and its relationship to the number of solutions of a quadratic equation etc. There comes a point when you must go back to your classnotes or textbook and review some of this material. Also, there are no simple recipes to problem solving ....
• Jan 14th 2010, 03:40 PM
Prove It
Quote:

Originally Posted by mr fantastic
Unfortunately, this advice needs to be taken with a grain of salt. However, I'm not going to open up the can of worms by saying that sometimes a factorised form or a turning point form might be more desirable than the standard form.

@OP: You have said in a number of threads that you are bad at maths etc. That's no excuse for not investing the necessary time and effort so that you have a basic understanding of things like the discriminant and its relationship to the number of solutions of a quadratic equation etc. There comes a point when you must go back to your classnotes or textbook and review some of this material. Also, there are no simple recipes to problem solving ....

I only meant, write it as $\displaystyle ax^2 + bx + c$ as opposed to $\displaystyle ax^2 - bx + c$.
• Jan 14th 2010, 03:43 PM
Bacterius
Why would you want to write it as $\displaystyle ax^2 - bx + c$ ? It would just be confusing, and since the discriminant heavily relies on the sign of $\displaystyle \Delta$, a little sign error could be fatal.
• Jan 15th 2010, 03:04 AM
Punch
Indeed, I have made quite a number of mistakes because of the + and - signs and they gave me answers that were world apart...

So I was confused where I made the mistake and hence decided to confirm on whether to write it as $\displaystyle ax^2 - bx + c$ or $\displaystyle ax^2 + bx + c$
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last