1. ## Difficult quadratic equation problems

Hi,

How do i solve these?

1)
If $\alpha$ and $\beta$ are the roots of the equation $ax^2 + bx + c$ = 0, then the equation whose roots are $\alpha + (1/\beta)$ and $\beta + (1/\alpha)$, is?

2)
If the equations $2x^2 + 3x + 5\gamma = 0$ and $x^2 + 2x + 3\gamma = 0$ have a common root, then $\gamma =$ ?

3)
If the difference of the roots of the equations $x^2 + bx + c = 0$ be 1, then:

(a) $b^2-4c-1 = 0$
(b) $b^2 - 4c = 0$
(c) $b^2 - 4c + 1 = 0$
(d) b $^2 + 4c - 1 = 0$

2. Hello,

Let's use brute force... for the first question :

If $\alpha,\beta$ are the roots, then $\alpha+\beta=-\frac ba$ and $\alpha\beta=\frac ca$

Let $x^2+\mu x+\nu=0$ be the equation which roots are $\alpha+\frac 1\beta,\beta+\frac 1\alpha$

Then $-\mu=\alpha+\tfrac 1\beta+\beta+\tfrac 1\alpha=(\alpha+\beta)+\tfrac{\alpha+\beta}{\alpha \beta}=-\tfrac ba-\tfrac bc$
So $\mu=\tfrac ba+\tfrac bc$

And $\nu=\left(\alpha+\tfrac 1\beta\right)\cdot\left(\beta+\tfrac 1\alpha\right)=\alpha\beta+\tfrac{1}{\alpha\beta}+ 2=\tfrac ca+\tfrac ac+2$

So the equation you're looking for is $x^2+\left(\tfrac ba+\tfrac bc\right)x+\tfrac ca+\tfrac ac+2=0 \Leftrightarrow acx^2+b(a+c)x+(a+c)^2=0$

3. For question 2)...

There exists $\delta$ such that :

$\begin{cases} 2\delta^2+3\delta+5\gamma=0 \\ \delta^2+2\delta+3\gamma=0\end{cases}$

You may be able to solve for $\gamma$ by yourself eh ?

For question 3)...

We know that the sum of the two roots $\alpha,\beta$ is -b and their product is c.
So we have $b^2=(\alpha+\beta)^2=\alpha^2+\beta^2+2\alpha\beta =\alpha^2+\beta^2+2c$

Hence $\alpha^2+\beta^2=b^2-2c$

But $1=(\alpha-\beta)^2=\alpha^2+\beta^2-2\alpha\beta=\alpha^2+\beta^2-2c=b^2-4c$

Finally, $b^2-4c-1=0$

4. Hello, saberteeth!

My approach to #1 is similar to Moo's . . . with a different answer.
. . Did I mess up?

1) If $\alpha$ and $\beta$ are the roots of the equation: $ax^2 + bx + c \:=\: 0$,
find the equation whose roots are: . $\alpha + \tfrac{1}{\beta}$ and $\beta + \tfrac{1}{\alpha}$ . . . . in terms of ${\color{blue}a,b,c}$

Since $\alpha,\:\beta$ are roots of: . $x^2 + \frac{b}{c}x + \frac{c}{a} \:=\:0$
. . then: . $\begin{array}{cccc}\alpha + \beta &=& \text{-}\dfrac{b}{a} & [1] \\ \\[-3mm] \alpha\beta &=& \dfrac{c}{a} & [2] \end{array}$

The sum of the two roots is:

. . $S \;=\;\left(\alpha + \frac{1}{\beta}\right) + \left(\beta + \frac{1}{\alpha}\right) \;=\; (\alpha + \beta) + \left(\frac{1}{\alpha} + \frac{1}{\beta}\right)$ . $=\;(\alpha + \beta) + \left(\frac{\alpha + \beta}{\alpha\beta}\right) \;=\;(\alpha+\beta)\left(1 + \frac{1}{\alpha\beta}\right)$

Substitute [1] and [2]: . $S \;=\;-\frac{b}{a}\left(1 + \frac{1}{\frac{c}{a}}\right) \;=\; -\frac{b(a+c)}{ac}$

. . Hence, the $x$-coefficient is: . $\frac{b(a+c)}{ac}$

The product of the two roots is:

. . $P \;=\;\left(\alpha + \frac{1}{\beta}\right)\left(\beta+\frac{1}{\alpha} \right) \;=\;\alpha\beta + 1 + 1 + \frac{1}{\alpha\beta}$

Substitute [2]: . $P \;=\;\frac{c}{a} + 2 + \frac{a}{c} \;=\;\frac{a^2 + 2ac + c^2}{ac}$

. . Hence, the constant term is: . $\frac{(a+c)^2}{ac}$

Therefore, the equation is: . $x^2 + \frac{b(a+c)}{ac}x + \frac{(a+c)^2}{ac} \:=\:0 \quad\Rightarrow\quad\boxed{ acx^2 + b(a+c)x + (a+c)^2 \:=\:0}$

5. Yep Soroban, a slight error in the signs and an incorrect factorisation

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