1. ## Help with finding roots of complicated quadratic

Hi

I'm having trouble finding the roots of the equation below. While I have the final answer, I'm not sure how to arrive at that step-by-step. Any help would be appreciated.

x^2 + (z ln z - zb)x - (z^2 * b * ln z) = 0

The final answer has 2 roots in terms of 3 variables: x, b, z

2. Originally Posted by ForgotMath
Hi

I'm having trouble finding the roots of the equation below. While I have the final answer, I'm not sure how to arrive at that step-by-step. Any help would be appreciated.

x^2 + (z ln z - zb)x - (z^2 * b * ln z) = 0

The final answer has 2 roots in terms of 3 variables: x, b, z

A = 1

B = (z ln z - zb)

C = - (z^2 * b * ln z)

Use the quadratic formula and simplify. Take a shovel and start digging.

3. Or, completing the square:

$x^2+\alpha{x}-\beta = \left(x+\frac{\alpha}{2}\right)^2-\left(\beta+\frac{\alpha^2}{4}\right).$

4. Originally Posted by mr fantastic
A = 1

B = (z ln z - zb)

C = - (z^2 * b * ln z)

Use the quadratic formula and simplify. Take a shovel and start digging.
I did try this before posting but couldn't get anywhere near the final answer.

-(z ln z - zb) +/- sqrt{z ln z -zb) * (z ln z -zb) - 4z^2*b*ln z} / 2
-z ln z + zb +/- sqrt{z^2 (ln z)^2 - z^2*b*ln z - z^2*b*ln z + z^2*b^2 - 4 z^2*b*ln z} / 2
-z ln z + zb +/- sqrt{z^2(ln z)^2 - 6z^2*b*ln z + z^2*b^2} / 2

(x - zb) (x + z ln z)

I don't see this simplifying to that, or at least I don't know how to do it.

Thanks

5. There is a mistake in your first line:

- 4z^2*b*ln z

should be:

+ 4z^2*b*ln z

Simplify further, and it factors nicely.

6. Originally Posted by rtblue
There is a mistake in your first line:

- 4z^2*b*ln z

should be:

+ 4z^2*b*ln z

Simplify further, and it factors nicely.
Thanks for pointing that out. However, the terms still do not cancel out so how do you get rid of the square root?

7. Originally Posted by ForgotMath
Thanks for pointing that out. However, the terms still do not cancel out so how do you get rid of the square root?
Let me simplify the problem a bit and you can return to the general case.

You are basically faced with the problem of simplifying an expression of the form
$(a - b)^2 + 4ab = a^2 - 2ab + b^2 + 4ab$

$(a - b)^2 + 4ab = a^2 + 2ab + b^2$

How can you simplify that?

-Dan

8. Originally Posted by topsquark
Let me simplify the problem a bit and you can return to the general case.

You are basically faced with the problem of simplifying an expression of the form
$(a - b)^2 + 4ab = a^2 - 2ab + b^2 + 4ab$

$(a - b)^2 + 4ab = a^2 + 2ab + b^2$

How can you simplify that?

-Dan
I'd like to see where the square root went...there is no square root in your equation and this is not similar to solving:
-z ln z + zb +/- sqrt{z^2(ln z)^2 + 2z^2*b*ln z + z^2*b^2} / 2

9. Originally Posted by ForgotMath
I'd like to see where the square root went...there is no square root in your equation and this is not similar to solving:
-z ln z + zb +/- sqrt{z^2(ln z)^2 + 2z^2*b*ln z + z^2*b^2} / 2
Yes it is. Look at your discriminant. It simplifies in the same way as I pointed out.

-Dan

10. Originally Posted by topsquark
Yes it is. Look at your discriminant. It simplifies in the same way as I pointed out.

-Dan
Maybe you should elaborate so the other person can actually see what they are doing wrong. Such a reply isn't very helpful - its been obvious from my all my posts that I've already to simplify it. I'm probably missing something, fine, but to keep saying "this simplifies" won't help me find my mistake.

11. [QUOTE=ForgotMath;643563]I did try this before posting but couldn't get anywhere near the final answer.
All right. A further hint on how to simplify the discriminant:
$(z ln z -zb)^2 + 4z^2*b*ln z$

(Note that in my previous post a = z ln(z) and b = zb.)

$= z^2~ln^2(z) - 2z^2b~ln(z) + z^2b^2 + 4z^2b~ln(z)$

$= z^2~ln^2(z) + 2z^2b~ln(z) + z^2b^2$

This is of the form $a^2 + 2ab + b^2$. So how do you simplify this further?

-Dan