1. ## Sum and Product of quadratic equations

Im very new to further maths and answered the following question -

The quadratic equation 3x²-9x+1=0 has roots α and β. Find a quadratic equation with integer coefficients and with roots 3/α , 3/β

This is how I answered it -

αβ = 1/3

α+β = 9/3 = 3

new sum = 3/α + 3/β

new sum = (3β + 3α) / (1/3)

new sum = 9 / (1/3) = 27

new product = 3/α * 3/β

new product = 9/αβ

= 9 / (1/3) = 27

x² - 27x + 27 = 0

Worked out right but now I'm stuck with this one. Its the same equation but with different roots -

The quadratic equation 3x²-9x+1=0 has roots α and β. Find a quadratic equation with integer coefficients and with roots 2α , 2β

I have started it out like this but getting the wrong answer -

αβ = 1/3

α + β = 9/3 = 3

new sum = 2α + 2β

new sum = 2(α+β)

= 2 * 3 = 6

However this is wrong :|

Could anybody give me any tips on how i answered the first question and help with the second question please

thanks

2. Hi djmccabie,

Both questions are similar in that the their root's relation to the previous roots are the same and so using a substitution may be easier. For example in question 2 we have $3x^2-9x+1=0$ which has roots $\alpha$ and $\beta$ . Find the quadratic equation with roots $2\alpha$ and $2\beta$ . Since the relation to the previous roots are the same we can define $w=2x$ (Note $x$ is the root of the first equation and $w$ is the root of the equation we are required to find) . This implies that $x=\frac{w}{2}$ . Subbing this into the first equation gives $3\left (\frac{w}{2}\right)^2-9\left (\frac{w}{2}\right)+1=\frac{3w^2}{4}-\frac{9w}{2}+1=3w^2-18w+4=0$ and thus, $3w^2-18w+4=0$ is the required equation.

In the first example a similar technique can be used except we define $w=\frac{3}{x}$

Now you started off right with the technique you chose in question 2. I think the reason you think it is wrong is because you get 6 and this isn't the middle term...... However you have not followed through. If you had you would notice you are right. We have $3x^2-9x+1=0$ . As you have already stated.....

$\sum\alpha=3$
$\alpha\beta=\frac{1}{3}$

Now the for the new equation we have,

$\sum \alpha=2\alpha+2\beta=2(\alpha+\beta)=6$
$\alpha\beta=2\alpha\cdot 2\beta = 4\alpha\beta = \frac{4}{3}$

So following on from that, $z^2-\left(\sum\alpha \right)z+\alpha\beta=z^2-6z+\frac{4}{3}=0$ . Thus our equation is $3z^2-18z+4=0$ .

Hope this helps.

3. Hi thanks for the reply! It did help a lot however i have a questions. As we havent been taught to sub in these equations im unsure how you got would it be possible to explain this further please Its ok if you dont have time you really helped me with my method

4. OK i have a new problem lol.

Using the same eqution again only this time the roots are a-2 , b-2

i managed to get the sum but cant find the product.

this is how far i get

new product = (a-2)(b-2)

new product = ab - 2a - 2b - 4

new product = ab - 2(a-b) - 4

new product = 1/3 - 2(a-b) - 4

Could someone please explain if my method is correct and how to solve the rest of the equation

thanks again

5. just spotted the silly mistake! that's if anyone can actually see this thread :/

6. Sorry I was at college.....

Originally Posted by djmccabie
Hi thanks for the reply! It did help a lot however i have a questions. As we havent been taught to sub in these equations im unsure how you got would it be possible to explain this further please Its ok if you dont have time you really helped me with my method
You will probably get taught this method very soon. We have an equation $ax^2+bx+c=0$ with roots $\alpha$ and $\beta$ . We know that the roots must satisfy the equation and hence we can say $a\alpha^2+b\alpha+c=0$ and $a\beta^2+b\beta+c=0$ so in other words we can say that the equation is satisfied when either $x=\alpha$ or $x=\beta$ . Now we want to find an equation, $dw^2+ew+f=0$ which has roots $2\alpha$ and $2\beta$ . Since these roots satisfy the equation we can say the equation is satisfied when either $w=2\alpha$ or $w=2\beta$ . Since $x=\alpha$ or $x=\beta$ we can conclude that $w=2x$ .