Quadratic equation with imaginary roots

**Stuck Man** · January 4th 2010, 06:02 AM

I have done part i of the attached question. Do I have to understand complex numbers for part ii?

**doomgaze** · January 4th 2010, 06:04 AM

Originally Posted by Stuck Man

I have done part i of the attached question. Do I have to understand complex numbers for part ii?

I don't see anything.

Try to post the question here or take a picture of it?

**Stuck Man** · January 4th 2010, 06:05 AM

**e^(i*pi)** · January 4th 2010, 06:09 AM

Originally Posted by Stuck Man

$\alpha = \frac{-b+\sqrt{b^2-4ac}}{2a}$

$\beta = \frac{-b-\sqrt{b^2-4ac}}{2a}$

**Stuck Man** · January 4th 2010, 06:18 AM

b^2-4ac is -23. I don't know how to get a square root of a negative so thats why I'm asking if this requires a knowledge of complex numbers.

**e^(i*pi)** · January 4th 2010, 06:44 AM

Originally Posted by Stuck Man

b^2-4ac is -23. I don't know how to get a square root of a negative so thats why I'm asking if this requires a knowledge of complex numbers.

Ah right, my mistake.

Yes, complex number knowledge is necessary IMO. Perhaps someone better than me can explain a way around it?

**Stuck Man** · January 4th 2010, 06:49 AM

I've solved ii now. It takes quite a lot of writing. Square roots of -23 get squared or cancel out near the end.

**Grandad** · January 4th 2010, 07:22 AM

Hello Stuck Man

It looks as if you don't know about this important result:

If $\alpha$ and $\beta$ are the roots of the quadratic equation $ax^2+bx+c=0$ , then:

The sum of its roots, $\alpha + \beta = -\frac{b}{a}$

and

The product of its roots, $\alpha\beta = \frac{c}{a}$

(This is very easy to prove: expand $(x-\alpha)(x-\beta)$ and compare it to $x^2 + \frac{b}{a}x + \frac{c}{a}$ .)

Well then, if $\alpha$ and $\beta$ are the roots of $3x^2 +x + 2 = 0$ , we get:

$\alpha + \beta = -\frac{1}{3}$

and

$\alpha\beta = \frac{2}{3}$

So for part (i):

$\frac{1}{\alpha^2}+\frac{1}{\beta^2} =\frac{\alpha^2+\beta^2}{\alpha^2\beta^2}$

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

Now substitute in the values of $\alpha + \beta$ and $\alpha\beta$ , simplify and the job's done. (I get the answer $-\frac{11}{4}$ . Do you agree?)

****************

Part (ii)

We know the value of $\frac{1}{\alpha^2}+\frac{1}{\beta^2}$ and clearly $\frac{1}{\alpha^2}\times\frac{1}{\beta^2}= \frac{1}{(\alpha\beta)^2}= \frac94$ . So we can use the result in reverse to write down the new equation, knowing the sum and product of its roots; i.e.

$x^2 - \text{(sum of roots)}x+ \text{(product of roots)}=0$

which (if my arithmetic is correct) is:

$x^2 +\frac{11}{4}x+\frac94=0$

i.e. $4x^2+11x+9=0$

****************

Part (iii)

$\alpha$ is a root of $3x^2+x+2=0$ . So:

$3\alpha^2+\alpha+2=0$

$\Rightarrow 3\alpha^2 = -\alpha-2$ (1)

$\Rightarrow 9\alpha^4 =(-\alpha-2)^2$

$=\alpha^2 +4\alpha+4$

$\Rightarrow 27\alpha^4 = 3\alpha^2 +12\alpha+12$

$=-\alpha - 2 +12 \alpha +12$ from (1)

$= 11\alpha+10$

Grandad

**Stuck Man** · January 5th 2010, 01:51 AM

I did part i the same. For parts ii and iii I used the quadratic formula to get a value for alpha. I got the answers but it was far more work.

Thanks.

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