Please helppp :) (1 Viewer)

umm what · Feb 10, 2012

please help with (iii) , (c) and (d)

Thanks

IamBread · Feb 10, 2012

You're going to need to show us the rest of the question for us to do (iii)

C.

$$A double root at $ x = 1 $ means $ p(1) = 0 $ and $ p'(1) = 0$

$p(x) = 2x^3 - (2a + 1)x^2 + (2 + b)x - 1 \\\\ p'(x) = 6x^2 -2(2a+1)x + (2+b) \\\\ p'(1) = 6 - 4a - 2 + 2 + b = 0 \\\\ 6 - 4a + b = 0 \\\\ p(1) = 2 - 2a - 1 + 2 + b - 1 = 0 \\\\ b- 2a +2 = 0$

$$Now we have some 2 equations we can solve simultaneously. $\\\\ 6 - 4a + b = 0 $ $[1] $ and $ b- 2a +2 = 0 $ $[2]$

$[1]$ $ 6 - 4a + b = 0 \\\\ -2a = -\frac{6+b}{2} \\\\ $Sub into [2] $\\\\ b - \frac{6+b}{2} + 2 = 0 \\\\ b -2= 0, b =2$

$$Sub $ b= 10 $ into [1] $ \\\\6 - 4a + 2 = 0 \\\\a =2$

d.

$$Let the roots be $\alpha, \beta, \alpha + \beta$

$\alpha + \beta + \alpha + \beta = -a \\\\ 2\alpha + 2\beta = -a $ $ [1]\\\\ \alpha \times \beta \times( \alpha + \beta) = -c \\\\ \alpha^2\beta + \alpha \beta^2 = -c $ $ [2]\\\\ \alpha \times \beta + \alpha \times (\alpha + \beta) + \beta \times ( \alpha + \beta) = b \\\\ \alpha\beta + \alpha^2 + \alpha\beta + \beta\alpha + \beta^2 = b \\\\ \alpha^2 + \beta^2 + 3\alpha\beta = b $ $ [3]$

$$From [3] $ \alpha\beta = \frac{b - \alpha^2 -\beta^2}{3} \\\\ $From [2] \alpha\beta(\alpha + \beta) = -c \\\\ \frac{b - \alpha^2 -\beta^2}{3}(\alpha + \beta) = -c$

$$From [1] $ (\alpha + \beta) = -\frac{a}{2} \\\\ \frac{b - (\alpha^2 +\beta^2)}{3} \times -\frac{a}{2} = -c $ $ [4]\\\\ \alpha^2 +\beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$

$[4] $ $ \frac{b - ((\alpha + \beta)^2 - 2\alpha\beta)}{3} \times -\frac{a}{2} = -c $$

$\frac{b - (\frac{a^2}{4} - \frac{4c}{a})}{3} \times -\frac{a}{2} = -c \\\\ a(b - (\frac{a^2}{4} - \frac{4c}{a})) = 6c \\\\ ab -\frac{a^3}{4} + 4c = 6c \\\\ 4ab - a^3 = 8c \\\\ a^3 -4ab +8c = 0 $ as required.$$

umm what · Feb 10, 2012

Yeah sureee

lolz

someth1ng · Feb 10, 2012

Okay, for (c)
You know there's a double root at x=1 so P(x)=?(x-1)(x-1)(x-gamma)

alpha*beta*gamma=-d/a=1/2
You know two roots are one and 1*1=1 so the last root is obviously 1/2.

Hence, P(x)=?(x-1)(x-1)(x-1/2)

Now, When you multiply -1,-1 and -1/2, it becomes -1/2 which is not -1.

.`. ?=2

Hence P(x)=2(x-1)(x-1)(x-1/2)

From here, you just solve by expanding then equating which I will leave for you to do.

someth1ng · Feb 10, 2012

IamBread, why can't I see a LaTeX button, the "f(x)" button or w/e button it was isn't here? :O

umm what · Feb 10, 2012

ohh thanks matee!! can u do (d) for meee plzzzzzzzzzz

IamBread · Feb 10, 2012

Wow d.... last time i do math on latex, going to do it on paper first from now on. Was such a good feeling when I came up with the right answer

IamBread · Feb 10, 2012

someth1ng said:
IamBread, why can't I see a LaTeX button, the "f(x)" button or w/e button it was isn't here? :O

You have to do it all by text, http://community.boredofstudies.org/showthread.php?t=234259 should help get you started.

umm what · Feb 10, 2012

ur a legenddd !!! LOL

THANKSSSS A LOTTT!!!!

SpiralFlex · Feb 10, 2012

Let the roots be alpha, beta, alpha + beta.

$\alpha + \beta = -\frac{a}{2}...(1)$

$(\alpha + \beta)^2+\alpha \beta=b...(2)$

$(\alpha+\beta)\alpha \beta = -c....(3)$

1 into 3,

$\alpha \beta=\frac{2c}{a}...(4)$

1, 4 into 2.

$(\frac{a}{2})^2+\frac{2c}{a}=b$

$a^3-4ab+8c^2=0$

someth1ng · Feb 10, 2012

There used to be a button for LaTeX, like a whole GUI - pretty much...

umm what · Feb 10, 2012

thnxxx

SpiralFlex · Feb 10, 2012

Why da long algaebra Iambreadz?

IamBread · Feb 10, 2012

SpiralFlex said:
Why da long algaebra Iambreadz?

Cos I was trying to do it in my head lol, next time I use paper first...

Timske · Feb 10, 2012

maths is so beautiful

SpiralFlex · Feb 10, 2012

Timske said:
maths is so beautiful

English is better.

IamBread · Feb 10, 2012

SpiralFlex said:
English is better.

Lies!

someth1ng · Feb 10, 2012

IamBread said:
Lies!

This is not a lie.

funnytomato · Feb 10, 2012

@OP, read the comment from IamBread, the very 1st line

Carrotsticks · Feb 10, 2012

Perhaps I am going blind, but (d) and (e) look pretty similar to me...

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