1. ## Roots of quadratic equation

The roots of the quadratic equation $\displaystyle 16x^2 + 7x + 4 = 0$ are $\displaystyle \alpha^2$ and $\displaystyle \beta^2$. Find two distinct quadratic equations whose roots are $\displaystyle \alpha$ and $\displaystyle \beta$.

My steps:

Sum of roots, $\displaystyle \alpha^2 + \beta^2 = \frac{-7}{16}$
Product of roots, $\displaystyle \alpha^2\beta^2 = \frac{1}{4}$

Sum of new roots, $\displaystyle \alpha + \beta = ... = \sqrt{\frac{-7}{16} + 2\sqrt\frac{1}{4}} = \pm\frac{3}{4}$
Product of new roots, $\displaystyle \alpha\beta = ... = \sqrt{\frac{1}{4}} = \pm\frac{1}{2}$

But it turns out I'm partially correct, because while there is a $\displaystyle \pm$ before $\displaystyle \frac{3}{4}$, there is not supposed to be a $\displaystyle \pm$ before $\displaystyle \frac{1}{2}$.

After staring at it for a couple minutes I'm guessing that since the original equation has a positive c/a, hence my new product of roots must be positive too. Am I right?

2. Originally Posted by fterh
The roots of the quadratic equation $\displaystyle 16x^2 + 7x + 4 = 0$ are $\displaystyle \alpha^2$ and $\displaystyle \beta^2$. Find two distinct quadratic equations whose roots are $\displaystyle \alpha$ and $\displaystyle \beta$.

My steps:

Sum of roots, $\displaystyle \alpha^2 + \beta^2 = \frac{-7}{16}$
Product of roots, $\displaystyle \alpha^2\beta^2 = \frac{1}{4}$

Sum of new roots, $\displaystyle \alpha + \beta = ... = \sqrt{\frac{-7}{16} + 2\sqrt\frac{1}{4}} = \pm\frac{3}{4}$
Product of new roots, $\displaystyle \alpha\beta = ... = \sqrt{\frac{1}{4}} = \pm\frac{1}{2}$

But it turns out I'm partially correct, because while there is a $\displaystyle \pm$ before $\displaystyle \frac{3}{4}$, there is not supposed to be a $\displaystyle \pm$ before $\displaystyle \frac{1}{2}$.

After staring at it for a couple minutes I'm guessing that since the original equation has a positive c/a, hence my new product of roots must be positive too. Am I right?
Dear fterh,

$\displaystyle \alpha+\beta=\sqrt{(\alpha+\beta)^2}=\sqrt{\alpha^ 2+\beta^2+2\alpha\beta}=\sqrt{-\frac{7}{16}+2\left(\frac{1}{2}\right)}=\frac{3}{4 }~not~\pm\frac{3}{4}$

3. Originally Posted by Sudharaka
Dear fterh,

$\displaystyle \alpha+\beta=\sqrt{(\alpha+\beta)^2}=\sqrt{\alpha^ 2+\beta^2+2\alpha\beta}=\sqrt{-\frac{7}{16}+2\left(\frac{1}{2}\right)}=\frac{3}{4 }~not~\pm\frac{3}{4}$
But the official answer in the book requires $\displaystyle \pm\frac{3}{4}$ to arrive at, and it makes sense, because squaring both $\displaystyle \frac{3}{4}$ and $\displaystyle \frac{-3}{4}$ gets the same answer.

4. Originally Posted by fterh
But the official answer in the book requires $\displaystyle \pm\frac{3}{4}$ to arrive at, and it makes sense, because squaring both $\displaystyle \frac{3}{4}$ and $\displaystyle \frac{-3}{4}$ gets the same answer.
Dear fterh,

When talking about the squareroot of a number we normally refer to the principle square root(the positive squareroot). Refer: Square root - Wikipedia, the free encyclopedia

The question says to find two distinct quadratic equations. So, $\displaystyle \alpha+\beta=\frac{3}{4}~and~\alpha\beta=\frac{1}{ 2}$ gives one quadratic equation. Whereas $\displaystyle \alpha+\beta=\frac{3}{4}~and~\alpha\beta=-\frac{1}{2}$ gives the other quadratic equation.

Hope this helps.

5. Originally Posted by Sudharaka
Dear fterh,

When talking about the squareroot of a number we normally refer to the principle square root(the positive squareroot). Refer: Square root - Wikipedia, the free encyclopedia

The question says to find two distinct quadratic equations. So, $\displaystyle \alpha+\beta=\frac{3}{4}~and~\alpha\beta=\frac{1}{ 2}$ gives one quadratic equation. And $\displaystyle \alpha\beta=-\frac{1}{2}$ gives the other quadratic equation.

Hope this helps.
I've discussed this extensively with a friend, and got it already But thanks for your help anyway. Although $\displaystyle \alpha\beta$ cannot be $\displaystyle \frac{-1}{2}$, because if it were we'd have a problem finding the sum of new roots because we'd be trying to root a negative number then.