1. Roots of a quadratic equation .

Q : If a, b and c $\in$ IR , with a is not 0 , and the roots of the equation $ax^2+bx+c=0$ are real , show that the roots of $a^2y^2-(b^2-2ac)y+c^2=0$ are also real .
If the roots of the quadratic equation $ax^2+bx+c=0$ are $\alpha$ and $\beta$ , state the value of $\alpha+\beta$ and $\alpha\beta$ in terms of a ,b and c . Hence , find the roots of the second equation in terms of $\alpha$ and $\beta$ .

I don understand this part only --- hence , find the roots of the second equation ( the last sentence )

Q : If a, b and c $\in$ IR , with a is not 0 , and the roots of the equation $ax^2+bx+c=0$ are real , show that the roots of $a^2y^2-(b^2-2ac)y+c^2=0$ are also real .
If the roots of the quadratic equation $ax^2+bx+c=0$ are $\alpha$ and $\beta$ , state the value of $\alpha+\beta$ and $\alpha\beta$ in terms of a ,b and c . Hence , find the roots of the second equation in terms of $\alpha$ and $\beta$ .

I don understand this part only --- hence , find the roots of the second equation ( the last sentence )
If the roots of $ax^2+bx+c=0$ are $\alpha$ and $\beta$ then

$ax^2+bx+c=0 \; \Rightarrow\; (x - \alpha)(x - \beta) = 0$ which on expanding and comparing gives

$(1)\;\;\;\alpha + \beta = - \frac{b}{a}\; \text{and}\; \alpha\, \beta = \frac{c}{a}$

The roots of your second equation are given by

$(2)\;\;\;r_1, r_2 = \frac{b^2-2ac \pm \sqrt{(b^2-2ac)^2 - 4a^2c^2}}{2a^2} = \frac{b^2-2ac \pm \sqrt{b^4-4ab^2c}}{2a^2}$

From (1), isolate $b\; \text{and}\;c$ and use these to eliminate them in (2). Simplify (the a's should cancel) and you get your result.

3. Re :

Thanks Danny , i understood fully what you did , but i encounter some problem in simplifying .

$\frac{b^2-2ac}{2a^2}+\frac{\sqrt{b^4-4ab^2c}}{2a^2}$

$=\frac{1}{2}(\frac{b}{a})^2-\frac{c}{a}+\sqrt{\frac{1}{4}(\frac{b}{a})^4}-\frac{c}{a}(\frac{b}{a})^2$

Substituting one in here , i seemed to make the equation a little complicated .

$\frac{1}{2}(\alpha+\beta)^2-\alpha\beta+\sqrt{\frac{1}{4}(\alpha+\beta)^4-\alpha\beta(\alpha+\beta)^2}$

which brings me nowhere . Or there is a simpler way to do this . Thanks

I think I understand what they want.
I don't know if I've found the shortest way, though . . .

If $a, b,c \in IR,\;a \neq 0,$ and the roots of the equation $ax^2+bx+c\:=\:0$ are real,
show that the roots of $a^2y^2-(b^2-2ac)y+c^2\:=\:0$ are also real.

If the roots of the quadratic equation $ax^2+bx+c\:=\:0$ are $\alpha, \beta$,
state the value of $\alpha+\beta$ and $\alpha\beta$ in terms of $a ,b, c.$

Hence, find the roots of the second equation in terms of $\alpha$ and $\beta.$
We already know that: . $\begin{Bmatrix}\alpha + \beta \:=\:-\dfrac{b}{a} \\ \alpha\beta \:=\:\dfrac{c}{a} \end{Bmatrix}$ . [1]

Using the Quadratic Formula on the second equation,

. . we have: . $y \;=\;\frac{(b^2-2ac) \pm b\sqrt{b^2-4ac}}{2a^2}$

We will now manipulate this expression beyond all recognition . . .

We have: . $y \;=\;\frac{b^2}{2a^2} - \frac{2ac}{2a^2 } \pm\frac{b\sqrt{b^2-4ac}}{2a^2}$ . $= \;\frac{1}{2}\!\cdot\!\frac{b^2}{a^2} - \frac{c}{a} \pm\frac{1}{2}\!\cdot\!\frac{b}{a}\!\cdot\!\frac{\ sqrt{b^2-4ac}}{a}$

. . . . . . . . $y\;= \;\frac{1}{2}\left(\frac{b}{a}\right)^2 - \left(\frac{c}{a}\right) \pm \frac{1}{2}\left(\frac{b}{a}\right) \sqrt{\left(\frac{b}{a}\right)^2 - 4\left(\frac{c}{a}\right)}$ . [2]

Substitute [1] into [2]:

. . $y \;=\;\frac{1}{2}(\text{-}\alpha-\beta)^2 - (\alpha\beta) \:\pm \:\frac{1}{2}(\text{-}\alpha-\beta)\sqrt{(\text{-}\alpha-\beta)^2 - 4(\alpha\beta)}$

. . $y \;=\;\frac{1}{2}(\alpha^2 + \beta^2) \:\mp\: \frac{1}{2}(\alpha + \beta)\sqrt{\alpha^2 - 2\alpha\beta + \beta^2}$

. . $y \;=\;\frac{1}{2}(\alpha^2+\beta^2) \:\mp\: \frac{1}{2}(\alpha + \beta)(\alpha - \beta)$

. . $y \;=\;\frac{1}{2}\bigg[(\alpha^2+\beta^2) \pm(\alpha^2-\beta^2)\bigg]$

Therefore: . $y \;=\;\alpha^2,\;\beta^2$

. . there is a more direct approach.
So far, I haven't found it . . . anyone? anyone?

Edit: Too fast for me, Danny!

5. Let $\gamma$ and $\delta$ be the roots of $a^2y^2-(b^2-2ac)y+c^2=0$

Then

$\begin{Bmatrix}\gamma + \delta \:=\:\dfrac{b^2-2ac}{a^2} \\ \gamma\delta \:=\:\dfrac{c^2}{a^2} \end{Bmatrix}$

$\dfrac{b^2-2ac}{a^2} = \left(\dfrac{b}{a}\right)^2 - 2 \:\dfrac{c}{a} = (\alpha + \beta)^2 - 2 \alpha \beta = \alpha^2 + \beta^2$

$\dfrac{c^2}{a^2} = \left(\dfrac{c}{a}\right)^2 = (\alpha \beta)^2 = \alpha^2 \beta^2$

Then

$\gamma + \delta = \alpha^2 + \beta^2$

$\gamma \delta = \alpha^2 \beta^2$

I am not sure this helps a lot

Originally Posted by running-gag
$\gamma + \delta = \alpha^2 + \beta^2$

$\gamma \delta = \alpha^2 \beta^2$

I am not sure this helps a lot
Isn't this exactly what Soroban was looking for? The roots are therefore $\alpha^2$ and $\beta^2$.

Isn't this exactly what Soroban was looking for? The roots are therefore $\alpha^2$ and $\beta^2$.
I was not completely sure that other pairs of $\gamma$ and $\delta$ could fit as well but no, you re right