Quadratic Equation Part 2

**cloud5** · June 24th 2009, 05:22 AM

How to do these?

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

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

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

Answer:
1. $x^2+3x-9=0$

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

3. 9pm= $2n^2$

**alexmahone** · June 24th 2009, 05:49 AM

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

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

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

= $\alpha+\beta$

= $-3$

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

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

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

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

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

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

= $9-18$

= $-9$

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

$x^2+3x-9=0$
-----------------------------------------------------------------------------------------------------

2. Let the roots of the original equation be $\alpha$ and $\beta$

$\alpha+\beta=-2a$

$\alpha\beta=-c^2$

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

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

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

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

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

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

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

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

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

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

$c^2x^2-2ax-1=0$
---------------------------------------------------------------------------------------------------------

3. Let the roots be $\alpha$ and $2\alpha$

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

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

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

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

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

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

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

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

$9pm=2n^2$

**Soroban** · June 24th 2009, 06:41 AM

Hello, cloud5!

We're expected to know this . . .

Given the quadratic equation: . $x^2 + px + q \:=\:0$ . . (The leading coefficient is 1.)

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

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

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

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

Answer: . $x^2+3x-9\:=\:0$

There is a dazzlingly clever method around all the expected algebra . . .

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

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

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

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

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

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

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

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

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

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

Find that equation.

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

The roots of the original equation are: . $\alpha\text{ and }\beta$

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

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

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

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

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

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

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

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

Care to try the third one youself?

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