Two quick maths questions-the discriminant (1 Viewer)

LoveHateSchool · Jul 31, 2011

I for the life of me can't figure out what I've done wrong in these two

1, Find the values of k for which 5+4x-x^2=k has equal roots.

2. If (2k+3)x^2-4kx+4=0 has equal roots, find k.

Rep and cookies for any helpful people.
I know that for roots to be equal, discriminant=0. I think I'm just messing up one of the steps in these two.

SpiralFlex · Jul 31, 2011

1. Rearrange,

$-x^2 + 4x + 5 - k = 0$

For equal roots $\Delta = 0$ .

$\Delta = 16 + 4(5-k) =0$

$16 + 20 - 4k = 0$

$k = 9$

2. For equal roots $\Delta = 0$ .

$\Delta = 16k^2 - 16(2k + 3) = 0$

$16k^2 - 32k - 48 = 0$

$16(k^2 - 2k - 3)=0$

$16(k+1)(k-3)=0$

$k= -1, 3$

Me want cookies! I FELL FOR THE TRAP! NOOOO.

Gigacube · Jul 31, 2011

:O Spi is back!

*Hands cookie*

SpiralFlex · Jul 31, 2011

Gigacube said:
:O Spi is back!

*Hands cookie*

I prefer my cookie to be human flavoured.

LoveHateSchool · Jul 31, 2011

Thank you Spi!

I have a question with the second one. In my book, it says it can also be 3, how does that work?

SpiralFlex · Jul 31, 2011

Mis-read question.

LoveHateSchool · Jul 31, 2011

^Thank you so much, I need to spread rep before giving to you again.

As a human cookie, I suggest you go for Stephanie Meyer. Doing the world a service I say

SpiralFlex · Jul 31, 2011

LoveHateSchool said:
^Thank you so much, I need to spread rep before giving to you again.

As a human cookie, I suggest you go for Stephanie Meyer. Doing the world a service I say

I prefer Edwards.

Gigacube · Jul 31, 2011

SpiralFlex said:
I prefer Edwards.

Vamps or Weres?

SpiralFlex · Jul 31, 2011

Both.

LoveHateSchool · Jul 31, 2011

I can't do another. I'm not sure if I rearranged it right to find the discriminant. It confuses me with all the k's and x's.

3. Find the values for k for which x^2 +(k+1)x=2k+1 have equal roots.

SpiralFlex · Jul 31, 2011

LoveHateSchool said:
I can't do another. I'm not sure if I rearranged it right to find the discriminant. It confuses me with all the k's and x's.

3. Find the values for k for which x^2 +(k+1)x=2k+1 have equal roots.

$x^2 +(k+1)x - (2k+1)=0$

For equal roots $\Delta = 0$

$\Delta = (k+1)^2 + 4(2k+1)=0$

$k^2 + 2k + 1 + 8k + 4 = 0$

$k^2 + 10k + 5 = 0$

$k = \frac{-10 \pm \sqrt{10^2 - 20}}{2}$

$k = \frac{-10 \pm 4\sqrt{5}}{2}$

$k = -5 \pm 2\sqrt{5}$

LoveHateSchool · Jul 31, 2011

^Spi, you can have my first born child for helping so much.

I have a question. The formula for discriminat is b^2-4ac, so therefore wouldn't it be -8 and -4. The answers in my book are k=-1 or -5 ?

SpiralFlex · Jul 31, 2011

Sure the question is $x^2 +(k+1)x=2k+1$ not $x^2 +(k-1)x=2k+1$ ?

$x^2 +(k-1)x=2k+1$ would make more sense.

If my suspicions are right then,

$x^2 +(k-1)x - (2k+1)=0$

For equal roots $\Delta = 0$

$\Delta = (k-1)^2 + 4(2k+1)=0$

$k^2 - 2k + 1 + 8k + 4 = 0$

$k^2 + 6k + 5 = 0$

$(k+1)(k+5)=0$

$\therefore k = -1, -5$

LoveHateSchool · Jul 31, 2011

Oh thank you Spi, I'll be sure to do some extra discriminant questions tomorrow arvo, I think I've gotten the idea now. Atm, I'm doing a revision sheet on lines (gradients, distance formula all that jazz).

SpiralFlex · Jul 31, 2011

Okay human.

Two quick maths questions-the discriminant (1 Viewer)

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