Another algebra challenge :( (1 Viewer)

coyazayo · Jan 18, 2016

Hi everyone just stuck on thesr Qs
Q.1: Given that x^2+x-1=0 find the exact value of x^5-5x

Q2.Given that x=(√2 +1)^1/3-(√2 -1)^1/3, find the exact value of x^3+3x-1

Q3.a, b, c,d are the positive real numbers when a>b>c>d. Given that bcd=a/9, acd=b, abd=9c/4 and abc=4d. Find the value of a+b-c+d

Q.4 Given that a+b+c=0, find the value of
a^2+1/bc + b^2+1/ca + c^2+1/ab

InteGrand · Jan 18, 2016

coyazayo said:
Hi everyone just stuck on thesr Qs
Q.1: Given that x^2+x-1=0 find the exact value of x^5-5x

Q2.Given that x=(√2 +1)^1/3-(√2 -1)^1/3, find the exact value of x^3+3x-1

Q3.a, b, c,d are the positive real numbers when a>b>c>d. Given that bcd=a/9, acd=b, abd=9c/4 and abc=4d. Find the value of a+b-c+d

Q.4 Given that a+b+c=0, find the value of
a^2+1/bc + b^2+1/ca + c^2+1/ab

$\underline{QUESTION 1}$

$\noindent The answer is $-3$.$

$\noindent Let $\Omega \equiv x^5 - 5x$. Note that we have $x^2 + x = 1 \Rightarrow x^5 + x^4 = x^3\Rightarrow x^5 = -x^4 +x^3$. So we have $\Omega \equiv x^5 - 5x = -x^4 + x^3 - 5x$ (replacing $x^5$ with $-x^4 +x^3$).$

$\noindent Now, note that $x^2 + x = 1 \Rightarrow x^4 + 2x^3 + x^2 = 1$ (squaring both sides), so $-x^4 = 2x^3 +x^2 -1$. Putting this into our last expression for $\Omega$, we have $\Omega = 2x^3 + x^2 -1 + x^3 - 5x \Rightarrow \Omega = 3x^3 + x^2 - 5x -1$.$

$\noindent Note that we are given $x^2 = 1-x$. Hence $x^3 = x-x^2$. Putting this into the last expression for $\Omega$, we have$

$\begin{align*}\Omega &= 3 \left(x-x^2\right) + x^2 - 5x -1 \\ &= -2x^2 -2x -1 \\ &= -2 \left(x^2 + x-1\right) -3 \\ &= 0-3 \\ &= -3.\end{align*}$

InteGrand · Jan 18, 2016

coyazayo said:
Hi everyone just stuck on thesr Qs
Q.1: Given that x^2+x-1=0 find the exact value of x^5-5x

Q2.Given that x=(√2 +1)^1/3-(√2 -1)^1/3, find the exact value of x^3+3x-1

Q3.a, b, c,d are the positive real numbers when a>b>c>d. Given that bcd=a/9, acd=b, abd=9c/4 and abc=4d. Find the value of a+b-c+d

Q.4 Given that a+b+c=0, find the value of
a^2+1/bc + b^2+1/ca + c^2+1/ab

$\underline{QUESTION 2}$

$\noindent The answer is 1. We have $x=\sqrt[3]{\sqrt{2}+1}-\sqrt[3]{\sqrt{2}-1}$. Note that $\left( \sqrt{2}+1\right)\left( \sqrt{2}-1\right)=2-1^2 = 1 \Rightarrow \sqrt{2}+1 = \frac{1}{\sqrt{2}-1}$. Let $\iota=\sqrt{2}+1$, so $\frac{1}{\iota}=\sqrt{2}-1$. Then we have$

$\begin{align*}x&= \sqrt[3]{\iota} - \sqrt[3]{\frac{1}{\iota}} \\ &= \sqrt[3]{\iota} - \frac{1}{ \sqrt[3]{\iota}} \\ \Rightarrow x^3 &= \iota - 3 \iota ^{\frac{2}{3}}\cdot \iota ^{-\frac{1}{3}}+ 3 \iota ^{\frac{1}{3}}\cdot \iota ^{-\frac{2}{3}} - \frac{1}{\iota} \quad (\text{expanding}) \\ &= \iota -3 \left(\sqrt[3]{\iota} - \frac{1}{ \sqrt[3]{\iota}} \right) - \frac{1}{\iota} \\ &= \iota -3x - \frac{1}{\iota} \\ \Rightarrow x^3 + 3x &= \iota - \frac{1}{\iota} \\ &= \sqrt{2}+1-\left(\sqrt{2}-1\right) \\ &= 2 \\ \Rightarrow x^3 + 3x -1 &= 1. \end{align*}$

Drsoccerball · Jan 18, 2016

InteGrand said:
$\underline{QUESTION 1}$

$\noindent The answer is $-3$.$

$\noindent Let $\Omega \equiv x^5 - 5x$. Note that we have $x^2 + x = 1 \Rightarrow x^5 + x^4 = x^3\Rightarrow x^5 = -x^4 +x^3$. So we have $\Omega \equiv x^5 - 5x = -x^4 + x^3 - 5x$ (replacing $x^5$ with $-x^4 +x^4$).$

$\noindent Now, note that $x^2 + x = 1 \Rightarrow x^4 + 2x^3 + x^2 = 1$ (squaring both sides), so $-x^4 = 2x^3 +x^2 -1$. Putting this into our last expression for $\Omega$, we have $\Omega = 2x^3 + x^2 -1 + x^3 - 5x \Rightarrow \Omega = 3x^3 + x^2 - 5x -1$.$

$\noindent Note that we are given $x^2 = 1-x$. Hence $x^3 = x-x^2$. Putting this into the last expression for $\Omega$, we have$

$\begin{align*}\Omega &= 3 \left(x-x^2\right) + x^2 - 5x -1 \\ &= -2x^2 -2x -1 \\ &= -2 \left(x^2 + x-1\right) -3 \\ &= 0-3 \\ &= -3.\end{align*}$

What's wrong with just completing the square AKA finding x and using substitution?

InteGrand · Jan 18, 2016

Drsoccerball said:
What's wrong with just completing the square AKA finding x and using substitution?

That's the cheap way out.

(Still valid.)

InteGrand · Jan 18, 2016

coyazayo said:
Hi everyone just stuck on thesr Qs
Q.1: Given that x^2+x-1=0 find the exact value of x^5-5x

Q2.Given that x=(√2 +1)^1/3-(√2 -1)^1/3, find the exact value of x^3+3x-1

Q3.a, b, c,d are the positive real numbers when a>b>c>d. Given that bcd=a/9, acd=b, abd=9c/4 and abc=4d. Find the value of a+b-c+d

Q.4 Given that a+b+c=0, find the value of
a^2+1/bc + b^2+1/ca + c^2+1/ab

$\underline{QUESTION 3}$

$\noindent Given:$

$$\left\{\begin{matrix}bcd=\frac{a}{9}\quad (1)\\acd=b\quad (2)\\abd = \frac{9c}{4} \quad (3)\\abc = 4d \quad (4)\\ \end{matrix}\right.$.$

$\noindent From (2), we have $cd=\frac{b}{a}$. So subbing in (1), we have$

$\begin{align*}b\cdot \frac{b}{a} &= \frac{a}{9}\Rightarrow b^2 &= \frac{a^2}{9} \\ \Rightarrow \boxed{b=\frac{a}{3}}.\end{align*}$

$\noindent From (3), $4=\frac{9c}{abd}\Rightarrow 4d = \frac{9c}{ab}$.$

$\noindent From (1), $9=\frac{a}{bcd}\Rightarrow 4d = \frac{9c}{ab}$. So in (4),$

$\begin{align*}\frac{9c}{ab} &= abc \\ \Rightarrow (ab)^2 &= 9 \\ \Rightarrow \boxed{ab = 3}. \end{align*}$

$\noindent Since $b=\frac{a}{3}$, we have $a\cdot\frac{a}{3}=3\Rightarrow a^2 = 9 \Rightarrow \boxed{a=3}$. Hence $b=\frac{3}{3}\Rightarrow \boxed{b=1}$. As $ab=3$, we have in (3)$

$\begin{align*}3d &= \frac{9c}{3} \\ \Rightarrow d&= \frac{3c}{4} \\ \Rightarrow \boxed{c = \frac{4}{3}d}.\end{align*}$

$\noindent Now in (2),$

$\begin{align*} acd &= b \Rightarrow 3 \cdot c \cdot \frac{3c}{4} &= 1 \\ \Rightarrow 9c^2 &= 4 \\ \Rightarrow \boxed{c = \frac{2}{3}}.\end{align*}$

$\noindent So $d=\frac{3}{4}c = \frac{3}{4}\cdot \frac{2}{3} \Rightarrow \boxed{d=\frac{1}{2}}$. Thus$

$\begin{align*}a+b+c+d&= 3 + 1 + \frac{2}{3} + \frac{1}{2} \\ &= \frac{31}{6}.\end{align*}$

$\noindent Edit: Realised it asked for $a+b-c+d$ instead. This is equal to $3 + 1 - \frac{2}{3} + \frac{1}{2}=\frac{23}{6}$.$

InteGrand · Jan 18, 2016

Another algebra challenge :( (1 Viewer)

coyazayo

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InteGrand

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InteGrand

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Drsoccerball

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InteGrand

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InteGrand

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