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

**Punch** · December 30th 2009, 06:48 PM

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!

**sym0110** · December 30th 2009, 08:08 PM

Originally Posted by Punch

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!

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:
$x+1=4x^2-2kx+k$ , rearranging gives:
$4x^2-(2k+1)x+(k-1)=0$ , now consider the discriminant of this quadratic:
$delta=(2k+1)^2-16(k-1)$
$delta=4((k-3/2)^2-7/4)$
implies that when k=3/2, delta<0, hence contradicts our assumption.

**Bacterius** · December 31st 2009, 12:01 AM

Hello sym0110,
for delta, you can use the \Delta command : $\Delta$

For the second question, you must show that for any $k$ , the equation $x + 1 = 4x^2 - 2kx + k$ has at least one real solution in $x$ . 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 : $4x^2 - 2kx + k - x - 1 = 0$ .

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

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

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

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

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

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

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

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

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

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 $a$ 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 :

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

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

**Punch** · December 31st 2009, 06:23 PM

Originally Posted by Bacterius

Hello sym0110,
for delta, you can use the \Delta command : $\Delta$

For the second question, you must show that for any $k$ , the equation $x + 1 = 4x^2 - 2kx + k$ has at least one real solution in $x$ . 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 : $4x^2 - 2kx + k - x - 1 = 0$ .

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

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

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

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

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

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

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

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

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

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 $a$ 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 :

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

And therefore we have proved that for any real $k$ , 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.

**Punch** · December 31st 2009, 06:26 PM

I do not understand why we have to prove $b^2-4ac>0$ .
And what the checking of discriminant is all about, neither do I know where $ \Delta ' = 12^2 - 4 \times 4 \times 17 = 144 - 272 = -128 \ (< 0) $ came from.

**Bacterius** · December 31st 2009, 06:28 PM

Then you should read this prior to messing with quadratic equations :

Quadratic equation - Wikipedia, the free encyclopedia

**Punch** · January 1st 2010, 03:44 AM

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.

**Boyrog** · January 7th 2010, 04:49 PM

@Integral
When k=3 the line also intersects the curve, the proof is down there.

@Punch the first part you didnt understand:

We want to prove that

, so

.

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.

Keep going :

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

.

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 4k²-12k+17, and there you can see it returns only positive values.

**HallsofIvy** · January 8th 2010, 03:21 AM

Originally Posted by Punch

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, $b^2-4ac < 0$

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

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

$ -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!

**Punch** · January 14th 2010, 01:15 AM

Originally Posted by Bacterius

Hello sym0110,
for delta, you can use the \Delta command : $\Delta$

For the second question, you must show that for any $k$ , the equation $x + 1 = 4x^2 - 2kx + k$ has at least one real solution in $x$ . 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 : $4x^2 - 2kx + k - x - 1 = 0$ .

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

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

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

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

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

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

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

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

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

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 $a$ 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 :

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

And therefore we have proved that for any real $k$ , 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 $ax^2-bx+c$ or $ax^2+bx+c$ or is it fine for both?

**Prove It** · January 14th 2010, 02:13 AM

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 $ax^2-bx+c$ or $ax^2+bx+c$ or is it fine for both?

Express every Quadratic equation as $ax^2 + bx + c$ .

**mr fantastic** · January 14th 2010, 03:44 AM

Originally Posted by Prove It

Express every Quadratic equation as $ax^2 + bx + c$ .

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 ....

**Prove It** · January 14th 2010, 03:40 PM

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 $ax^2 + bx + c$ as opposed to $ax^2 - bx + c$ .

**Bacterius** · January 14th 2010, 03:43 PM

Why would you want to write it as $ax^2 - bx + c$ ? It would just be confusing, and since the discriminant heavily relies on the sign of $\Delta$ , a little sign error could be fatal.

**Punch** · January 15th 2010, 03:04 AM

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 $ax^2 - bx + c$ or $ax^2 + bx + c$

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