1. ## Quadratic Equation Part 2

How to do these?

1. The roots of the quadratic equation $\displaystyle x^2+3x+1=0$are $\displaystyle \alpha$and$\displaystyle \beta$. Find the equation when the roots are $\displaystyle 2\alpha-\beta,2\beta-\alpha$

2. The roots of an equation are the reciprocals of the roots of the equation$\displaystyle x^2+2ax-c^2=0$. Find that equation.

3. If one of the roots of the equation $\displaystyle mx^2+nx+p=0$is twice the other root, find the relation between m,n and p.

1. $\displaystyle x^2+3x-9=0$

2. $\displaystyle c^2x^2-2ax-1=0$

3. 9pm=$\displaystyle 2n^2$

2. 1. From the first quadratic equation, $\displaystyle \alpha+\beta=-3$
$\displaystyle \alpha\beta=1$

Let the new roots be $\displaystyle \alpha'$ and $\displaystyle \beta'$

$\displaystyle \alpha'+\beta' = (2\alpha-\beta)+(2\beta-\alpha)$

= $\displaystyle \alpha+\beta$

= $\displaystyle -3$

$\displaystyle \alpha'\beta' = (2\alpha-\beta)(2\beta-\alpha)$

= $\displaystyle 4\alpha\beta-2{\beta}^2-2{\alpha}^2+\alpha\beta$

= $\displaystyle 5\alpha\beta-2({\alpha}^2+{\beta}^2)$

= $\displaystyle 5\alpha\beta-2({\alpha}^2+{\beta}^2+2{\alpha}{\beta})+4{\alpha} {\beta}$

= $\displaystyle 9\alpha\beta-2(\alpha+\beta)^2$

= $\displaystyle 9-2(-3)^2$

= $\displaystyle 9-18$

= $\displaystyle -9$

The new quadratic equation can be written as $\displaystyle x^2-x(\alpha'+\beta')+\alpha'\beta'=0$
$\displaystyle x^2-x(-3)+(-9)=0$

$\displaystyle x^2+3x-9=0$
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2. Let the roots of the original equation be $\displaystyle \alpha$ and $\displaystyle \beta$

$\displaystyle \alpha+\beta=-2a$

$\displaystyle \alpha\beta=-c^2$

Let the new roots be $\displaystyle \alpha'$ and $\displaystyle \beta'$

$\displaystyle \alpha'+\beta'=\frac{1}{\alpha}+\frac{1}{\beta}$

$\displaystyle \alpha'+\beta'=\frac{{\alpha}+{\beta}}{\alpha\beta }$

$\displaystyle \alpha'+\beta'=\frac{-2a}{-c^2}$

$\displaystyle \alpha'+\beta'=\frac{2a}{c^2}$

$\displaystyle \alpha'\beta'=\frac{1}{\alpha}\frac{1}{\beta}$

$\displaystyle \alpha'\beta'=\frac{-1}{c^2}$

Let the new roots be $\displaystyle \alpha'$ and $\displaystyle \beta'$

The new quadratic equation can be written as $\displaystyle x^2-x(\alpha'+\beta')+\alpha'\beta'=0$

$\displaystyle x^2-x(\frac{2a}{c^2})+\frac{-1}{c^2}=0$

$\displaystyle c^2x^2-2ax-1=0$
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3. Let the roots be $\displaystyle \alpha$ and $\displaystyle 2\alpha$

$\displaystyle \alpha+2\alpha=-\frac{n}{m}$

$\displaystyle 3\alpha=-\frac{n}{m}$

$\displaystyle 9{\alpha}^2=\frac{n^2}{m^2}$ -------------(1)

$\displaystyle \alpha*2\alpha=\frac{p}{m}$

$\displaystyle 2{\alpha}^2=\frac{p}{m}$ ------------ (2)

Equating $\displaystyle {\alpha}^2$ from equations (1) and (2),

$\displaystyle \frac{n^2}{9m^2}=\frac{p}{2m}$

$\displaystyle \frac{n^2}{9m}=\frac{p}{2}$

$\displaystyle 9pm=2n^2$

3. Hello, cloud5!

We're expected to know this . . .

Given the quadratic equation: .$\displaystyle x^2 + px + q \:=\:0$ . .

. . the roots $\displaystyle \alpha\text{ and }\beta$ are such that: .$\displaystyle \begin{Bmatrix}\alpha + \beta &=& -p \\ \alpha\beta &=& q \end{Bmatrix}$

The sum of the roots is the negative of the $\displaystyle x$-coefficient.
The product of the roots is the constant term.

1. The roots of the quadratic equation $\displaystyle x^2+3x+1\:=\:0$ are $\displaystyle \alpha$ and $\displaystyle \beta$.

Find the equation when the roots are: $\displaystyle 2\alpha-\beta,\;2\beta-\alpha$

Answer: .$\displaystyle x^2+3x-9\:=\:0$
There is a dazzlingly clever method around all the expected algebra . . .

The new equation has roots: .$\displaystyle 2\alpha-\beta\text{ and }2\beta-\alpha$

. . Its $\displaystyle x$-coefficient will be: .$\displaystyle p \;=\;-\bigg[(2\alpha-\beta) + (2\beta - \alpha)\bigg] \;=\;-(\alpha + \beta)$ .(a)

. . The constant term will be: .$\displaystyle q \;=\;-(2\alpha-\beta)(2\beta-\alpha) \:=\:-2(\alpha^2 + \beta^2) + 5(\alpha\beta)$ .(b)

From the original equation, we have: .$\displaystyle \begin{Bmatrix}\alpha + \beta &=& \text{-}3 & {\color{blue}[1]}\; \\ \alpha\beta &=& 1 & {\color{blue}[2]\;} \end{Bmatrix}$

Substitute [1] into (a): .$\displaystyle p \;=\;-(-3) \quad\Rightarrow\quad \boxed{p \:=\:3}$

$\displaystyle \text{Square }{\color{blue}[1]}\!:\quad (\alpha + \beta)^2 \:=\:(-3)^2 \quad\Rightarrow\quad (\alpha^2 + \beta^2) + 2\!\!\underbrace{(\alpha\beta)}_{\text{This is 1}} \:=\:9$

. . Hence: .$\displaystyle (\alpha^2+\beta^2) + 2 \:=\:9 \quad\Rightarrow\quad \alpha^2 + \beta^2 \:=\:7\;\;{\color{blue}[3]}$

Substitute [2] and [3] into (b): .$\displaystyle q \;=\;-2(7) + 5(1) \quad\Rightarrow\quad \boxed{q\;=\;-9}$

Therefore, the new equation is: .$\displaystyle \boxed{x^2 + 3x - 9 \:=\:0}$

2. The roots of an equation are the reciprocals of the roots of: .$\displaystyle x^2+2ax-c^2\:=\:0$

Find that equation.

Answer: .$\displaystyle c^2x^2 - 2ax - 1 \:=\:0$
The roots of the original equation are: .$\displaystyle \alpha\text{ and }\beta$

. . So we have: .$\displaystyle \begin{Bmatrix}\alpha + \beta &=& \text{-}2a & {\color{blue}[1]} \\ \alpha\beta &=& \text{-}c^2 & {\color{blue}[2]} \end{Bmatrix}$

The roots of the new equation are: .$\displaystyle \frac{1}{\alpha}\text{ and }\frac{1}{\beta}$

Its x-coefficient will be: .$\displaystyle p \;=\;-\left(\frac{1}{\alpha} + \frac{1}{\beta}\right) \;=\;-\left(\frac{\alpha+\beta}{\alpha\beta}\right)$

. . Substitute [1] and [2]: .$\displaystyle p \:=\:-\left(\frac{\text{-}2a}{\text{-}c^2}\right) \quad\Rightarrow\quad\boxed{ p \;=\;-\frac{2a}{c^2}}$

Its constant term will be: .$\displaystyle q \;=\;\frac{1}{\alpha}\!\cdot\!\frac{1}{\beta} \;=\;\frac{1}{\alpha\beta}$

. . Substitute [2]: .$\displaystyle q \;=\;\frac{1}{\text{-}c^2} \quad\Rightarrow\quad\boxed{ q \;=\;-\frac{1}{c^2}}$

The new equation is: .$\displaystyle x^2 - \frac{2a}{c^2}x - \frac{1}{c^2} \;=\;0$

Multiply by $\displaystyle c^2\!:\quad \boxed{c^2x^2 - 2ax - 1 \;=\;0}$

Care to try the third one youself?