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

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

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

4) Simplify $\frac{1}{\sqrt{x+1}}+\sqrt{x}$, 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 $\alpha$ and $\beta$ are the roots of the equation: $x^2 - 5x + 4 \:=\: 0$, find:

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

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

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

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

Quote:

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

Quote:

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

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

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

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

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

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

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

thanks for the help!

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