# some simple questions.from my exam(roots ,partial fractions ..)

• Jun 12th 2009, 10:26 AM
llkkjj24
some simple questions.from my exam(roots ,partial fractions ..)
hi guys i had an exam today and just wanted you guys to give me the correct answer for it so i can see how i went.
i'm sure its really easy for you guys(Wink)

1) if $\displaystyle \alpha$and$\displaystyle \beta$ are the roots of the equation x^2 - 5x + 4 = 0 find
i) $\displaystyle \alpha^2 + \beta^2$ ii) $\displaystyle \alpha^2 \beta^2$ iii) the equation with roots $\displaystyle \alpha^2$ and $\displaystyle \beta^2$

2) the curve$\displaystyle f(x) = ax^2 + bx + c$ has factors$\displaystyle (x + 1)$ and$\displaystyle (x + 3)$and cuts the y-axis at $\displaystyle y = 8$ . Find the values of a,b,c

3) Express in partial fractions $\displaystyle \frac{2}{(x-2)(2x^2+1)}$ in partial fractions.

4) Simplify $\displaystyle \frac{1}{\sqrt{x+1}}+\sqrt{x}$, rationalising the denominator.

thanks for the help!
• Jun 12th 2009, 11:09 AM
dhiab
• Jun 12th 2009, 11:18 AM
Soroban
Hello, llkkjj24!

Quote:

1) if $\displaystyle \alpha$ and $\displaystyle \beta$ are the roots of the equation: $\displaystyle x^2 - 5x + 4 \:=\: 0$, find:

$\displaystyle (i)\;\;\alpha^2 + \beta^2$

We know that: .$\displaystyle \begin{array}{cc}\alpha + \beta \:=\:5 & [1] \\ \alpha\!\cdot\!\beta \:=\:4 & [2] \end{array}$

$\displaystyle \text{Square [1]: }\;(\alpha + \beta)^2 \:=\:5^2 \quad\Rightarrow\quad \alpha^2 + 2(\alpha\beta) + \beta^2 \:=\:25$
. . . . . . . . . . . . . . . . . . . . . . . . . .$\displaystyle \uparrow$
. . . . . . . . . . . . . . . . . . . . . . . . $\displaystyle ^{\text{This is 4}}$

And we have: .$\displaystyle \alpha^2 + 8 + \beta^2 \:=\:25 \quad\Rightarrow\quad \alpha^2 + \beta^2 \:=\:17$

Quote:

$\displaystyle (ii)\;\;\alpha^2 \beta^2$
Since $\displaystyle \alpha\beta \:=\:4$, then: .$\displaystyle \alpha^2\beta^2 \:=\:16$

Quote:

$\displaystyle (iii)$ the equation with roots $\displaystyle \alpha^2$ and $\displaystyle \beta^2$
The equation is: .$\displaystyle (x - \alpha^2)(x - \beta^2) \:=\:0 \quad\Rightarrow\quad x^2 - \alpha^2x - \beta^2x + \alpha^2\beta^2\:=\:0$

. . $\displaystyle x^2 - \underbrace{(\alpha^2 + \beta^2)}x + \underbrace{\alpha^2\beta^2} \:=\:0$
. . . . . . . .$\displaystyle \uparrow$ . . . . . .$\displaystyle \uparrow$
. . . . . .$\displaystyle ^{\text{This is 17}}$ . . $\displaystyle ^{\text{This is 16}}$

Therefore: .$\displaystyle x^2 - 17x + 16 \;=\;0$

• Jun 12th 2009, 11:40 AM
dhiab
Quote:

Originally Posted by llkkjj24
hi guys i had an exam today and just wanted you guys to give me the correct answer for it so i can see how i went.
i'm sure its really easy for you guys(Wink)

1) if $\displaystyle \alpha$and$\displaystyle \beta$ are the roots of the equation x^2 - 5x + 4 = 0 find
i) $\displaystyle \alpha^2 + \beta^2$ ii) $\displaystyle \alpha^2 \beta^2$ iii) the equation with roots $\displaystyle \alpha^2$ and $\displaystyle \beta^2$

2) the curve$\displaystyle f(x) = ax^2 + bx + c$ has factors$\displaystyle (x + 1)$ and$\displaystyle (x + 3)$and cuts the y-axis at $\displaystyle y = 8$ . Find the values of a,b,c

3) Express in partial fractions $\displaystyle \frac{2}{(x-2)(2x^2+1)}$ in partial fractions.

4) Simplify $\displaystyle \frac{1}{\sqrt{x+1}}+\sqrt{x}$, rationalising the denominator.

thanks for the help!

HELLO : QUESTION 2
Ihave the system http://www.mathramz.com/xyz/latexren...a2bd748fe9.png
Continu......