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Maths Help? :(


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#1 Gmac

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Posted 07 December 2008 - 10:10 PM

1st post here so go easy on me laugh.gif

I'm really stuck with 4 questions. Any help would be great thanks smile.gif

#1) Factorise fully:

2x^2 + 5x^2 - 4x - 3



#2) For what value of "a" does the equation

ax^2+ 20x + 40 = 0

have equal roots?


#3) Show that the roots of the equation

(k-2)x^2 - (3k-2)x + 2k = 0

are real?




#4)

Attached Image: Maths.jpg

#2 Marcus

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Posted 07 December 2008 - 11:19 PM

Question 1)

first I think you've made a typo I think it is supposed to be f(x)=2x^{3}+5x^{2}-4-3, to find the first root is the most difficult, as it involves a bit of guesswork
you need to find a x value such that, f(x)=0, now x=1 works so (x-1) must be a factor then you need to use synthetic division
CODE
1 |  2  5  -4  -3
  |     2   7   3  
  _________________
     2  7   3   0


this tells us that f(x)=(x-1)(2x^{2}+7x+3)

now all we need to do is factorise that quadratic (if possible) now 2x^{2}+7x+3=(2x+1)(x+3)
so
2x^{3}+5x^{2}-4-3 in its fully factorised form is (x-1)(2x+1)(x+3)

Question 2)

now this question involves the discriminant, for equal roots b-4ac=0
so a=a, b=20, c=40


\begin{align*}
(20)^{2}-(4)(a)(40)&=0 \\
400-160a&=0 \\
400&=160a \\
a&=2.5 
\end{align*}

Question 3)

this is another discriminant question, but for real roots b-4ac >=0 or greater than or equal to zero
in this case a = (k-2), b= -(3k-2) c = 2k
(-(3k-2))^{2} - (4)(k-2)(2k)
=9k^{2}-12k+4-8k^{2} +16k
=k^{2}+4k+4
now this is the challenge, to try and show that this quadratic is positive or zero for all values of k
Let's consider its graph
You can tell that if we put in a positive value this quadratic will be positive, so at least some of it is above the x-axis
if you factorise it you get (k+2) which shows that its turning point is on the x-axis therefore it doesn't cross the x-axis.
Therefore, since part of the graph is above the x-axis, and it doesn't cross the x-axis (and the graph is continuous) therefore it is always above or equal to 0
Though this may be overkill compared to what you are expected to do

Question 4)
This is a nasty differentiation question for higher

f(x)=\frac{5x^{4}+6x^{2}-3}{2\sqrt{x}} //now we need to turn that 2\sqrt{x} into 2x^{-1/2} and multiply the top line by it
f(x)=(5x^{4})(2x^{-1/2})+(6x^{2})(2x^{-1/2})-3(2x^{-1/2}) //simplify using laws of indices
f(x)=10x^{7/2}+12x^{3/2}-6x^{-1/2}

then just differentiate like you would any other polynomial I'll let you do this yourself, if you struggle just post again and I'll write it out

Hope this helps, and I hope I haven't made any mistakes XD


=-=-=Marcus=-=-=

#3 Gmac

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Posted 08 December 2008 - 05:59 PM

QUOTE (Marcus @ Dec 7 2008, 11:19 PM) <{POST_SNAPBACK}>
Question 1)

first I think you've made a typo I think it is supposed to be f(x)=2x^{3}+5x^{2}-4-3, to find the first root is the most difficult, as it involves a bit of guesswork
you need to find a x value such that, f(x)=0, now x=1 works so (x-1) must be a factor then you need to use synthetic division
CODE
1 |  2  5  -4  -3
  |     2   7   3  
  _________________
     2  7   3   0


this tells us that f(x)=(x-1)(2x^{2}+7x+3)

now all we need to do is factorise that quadratic (if possible) now 2x^{2}+7x+3=(2x+1)(x+3)
so
2x^{3}+5x^{2}-4-3 in its fully factorised form is (x-1)(2x+1)(x+3)

Question 2)

now this question involves the discriminant, for equal roots b-4ac=0
so a=a, b=20, c=40


\begin{align*}
(20)^{2}-(4)(a)(40)&=0 \\
400-160a&=0 \\
400&=160a \\
a&=2.5 
\end{align*}

Question 3)

this is another discriminant question, but for real roots b-4ac >=0 or greater than or equal to zero
in this case a = (k-2), b= -(3k-2) c = 2k
(-(3k-2))^{2} - (4)(k-2)(2k)
=9k^{2}-12k+4-8k^{2} +16k
=k^{2}+4k+4
now this is the challenge, to try and show that this quadratic is positive or zero for all values of k
Let's consider its graph
You can tell that if we put in a positive value this quadratic will be positive, so at least some of it is above the x-axis
if you factorise it you get (k+2) which shows that its turning point is on the x-axis therefore it doesn't cross the x-axis.
Therefore, since part of the graph is above the x-axis, and it doesn't cross the x-axis (and the graph is continuous) therefore it is always above or equal to 0
Though this may be overkill compared to what you are expected to do

Question 4)
This is a nasty differentiation question for higher

f(x)=\frac{5x^{4}+6x^{2}-3}{2\sqrt{x}} //now we need to turn that 2\sqrt{x} into 2x^{-1/2} and multiply the top line by it
f(x)=(5x^{4})(2x^{-1/2})+(6x^{2})(2x^{-1/2})-3(2x^{-1/2}) //simplify using laws of indices
f(x)=10x^{7/2}+12x^{3/2}-6x^{-1/2}

then just differentiate like you would any other polynomial I'll let you do this yourself, if you struggle just post again and I'll write it out

Hope this helps, and I hope I haven't made any mistakes XD

Thanks Marcus

Really helped me understand

Owe you one mate! cool.gif





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