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

1. 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!

2. 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.

3. 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).

4. 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.

5. 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.

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

Quadratic equation - Wikipedia, the free encyclopedia

7. 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.

8. @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.

9. 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 "$\displaystyle -2k^2$".

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

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

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

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

ii) Stucked!

10. 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?

11. 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 $\displaystyle ax^2 + bx + c$.

12. 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 ....

13. 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 $\displaystyle ax^2 - bx + c$.

14. 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.

15. 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 $\displaystyle ax^2 + bx + c$

Page 1 of 2 12 Last