1. ## Help with a quadratic equation, that has a radical fraction

I hate fractions, but I don't usually get them wrong. I didn't really get this one wrong, but I'm not really sure of the complete process with this one.

I'm 22 and since I dropped out of high school, I plan on getting my GED, so I bought a college algebra book. The book is "McGraw Hill - College Algebra Demystified" and it seems like a pretty good book, but lacks a lot explanations. I have done quadratic equations before, but not radicals with fractions.

This is the problem: 3x^2+9x-2=0

I'm on the first chapter of this book and I'm completing the square and I don't have any problem there. The problem is with my answer and their extra step with the fraction.

I end up with: x = -3/2 + √35/12 and -3/2 - √35/12

The books answer is: x = -3/2 + √105/6 and -3/2 - √105/6

I understand that you get: 2√3 for 12. But why are the numerator and denominator both multiplied by 3?

I feel crazy asking this question, because I can see how it's done, but I need an explanation on the last part.

2. Originally Posted by panther22
I hate fractions, but I don't usually get them wrong. I didn't really get this one wrong, but I'm not really sure of the complete process with this one.

I'm 22 and since I dropped out of high school, I plan on getting my GED, so I bought a college algebra book. The book is "McGraw Hill - College Algebra Demystified" and it seems like a pretty good book, but lacks a lot explanations. I have done quadratic equations before, but not radicals with fractions.

This is the problem: 3x^2+9x-2=0

I'm on the first chapter of this book and I'm completing the square and I don't have any problem there. The problem is with my answer and their extra step with the fraction.

I end up with: x = -3/2 + √35/12 and -3/2 - √35/12

The books answer is: x = -3/2 + √105/6 and -3/2 - √105/6

I understand that you get: 2√3 for 12. But why are the numerator and denominator both multiplied by 3?

I feel crazy asking this question, because I can see how it's done, but I need an explanation on the last part.

$\displaystyle x=\frac{-9\pm\sqrt{9^2-4(-2)(3)}}{2\times3}$

$\displaystyle \Rightarrow x=\frac{-9\pm\sqrt{81+24}}{6}$

$\displaystyle x=\left\{\frac{-9+\sqrt{105}}{6},\frac{-9-\sqrt{105}}{6}\right\}$

$\displaystyle 3x^2+9x-2=0$

Start by dividing everything by the leading coefficient (3).

$\displaystyle \frac{3x^2+9x-2}{3}=\frac{0}{3}$

$\displaystyle x^2+3x-\frac{2}{3}=0$

add two thirds to both sides

$\displaystyle x^2+3x=\frac{2}{3}$

Complete the square by adding one half of the square of the linear coefficient to both sides

$\displaystyle x^2+3x+\frac{9}{4}=\frac{2}{3}+\frac{9}{4}$

Now that we have a trinomial square we factor

$\displaystyle (x+\frac{3}{2})^2=\frac{2}{3}+\frac{9}{4}$

simplifying and taking the square root

$\displaystyle x+\frac{3}{2}=\pm\sqrt{\frac{35}{12}}$

subtracting three halves to both sides

$\displaystyle x=\pm\sqrt{\frac{35}{12}}-\frac{3}{2}$

$\displaystyle x=\pm\frac{\sqrt{35}}{2\sqrt{3}}-\frac{3}{2}$

Now here's where you get stuck right? Well this part's easy. they multiplied $\displaystyle \frac{3}{2}\cdot\frac{{3}}{{3}}$ because it does two things. It finds the common denominator AND rationalizes the denominator at the same time. cross cancelling.

EG, if i cross cancel $\displaystyle \frac{1}{\sqrt{3}}\cdot\frac{3}{2}$
what happens? Well, root 3 will go into itself one time, and root 3 will go into 3.....................................root 3 times! So our example becomes

$\displaystyle \frac{1}{1}\cdot\frac{\sqrt{3}}{2}=\frac{\sqrt{3}} {2}$

Can you take it from here?

4. Originally Posted by VonNemo19
$\displaystyle x=\pm\sqrt{\frac{35}{12}}-\frac{3}{2}$

$\displaystyle x=\pm\frac{\sqrt{35}}{2\sqrt{3}}-\frac{3}{2}$

Now here's where you get stuck right? Well this part's easy. they multiplied $\displaystyle \frac{3}{2}\cdot\frac{{3}}{{3}}$ because it does two things. It finds the common denominator AND rationalizes the denominator at the same time. cross cancelling.

EG, if i cross cancel $\displaystyle \frac{1}{\sqrt{3}}\cdot\frac{3}{2}$
what happens? Well, root 3 will go into itself one time, and root 3 will go into 3.....................................root 3 times! So our example becomes

$\displaystyle \frac{1}{1}\cdot\frac{\sqrt{3}}{2}=\frac{\sqrt{3}} {2}$

Can you take it from here?
This is the books example:
$\displaystyle x+\frac{3}{2}=\pm\sqrt{\frac{35}{12}}=\pm\frac{\sq rt{35}}{\sqrt{12}}=\pm\frac{\sqrt{35}}{\sqrt{4}\sq rt{3}}$

$\displaystyle x=\frac{3}{2}=\pm\frac{\sqrt{35}}{2\sqrt{3}}\cdot\ frac{\sqrt{3}}{\sqrt{3}}=\pm\frac{\sqrt{105}}{2\cd ot{3}}=\pm\frac{\sqrt{105}}{6}$

$\displaystyle x=-\frac{3}{2}\pm\frac{\sqrt{105}}{6}$

I know that $\displaystyle 2\sqrt{3}=12$, but they didn't explain why the numerator and denominator both were multiplied by 3. I could get the final answer with the quadratic formula, but since I'm completing the square, that extra step was done, but not explained. I can see what they did, but seeing it and being able to do all the problems, without the explanation is pointless. That's why I asked this question, because I've never had a radical in a fraction.

Thanks for the help. I just need a worksheet with about 50-100 problems exactly like this and I should have it drilled in my brain, because radicals in fractions are new to me.

5. They multiplied the top and bottom by $\displaystyle \sqrt{3}$ because it's not considered proper to have a surd in the denominator.

6. Originally Posted by Stroodle
They multiplied the top and bottom by $\displaystyle \sqrt{3}$ because it's not considered proper to have a surd in the denominator.
Well, I guess "surd" will be my word of the day. I looked at the definition on Wolfram Alpha's site and the article about it on wikipedia and that made it a lot clearer.

7. Originally Posted by panther22
I hate fractions, but I don't usually get them wrong. I didn't really get this one wrong, but I'm not really sure of the complete process with this one.

I'm 22 and since I dropped out of high school, I plan on getting my GED, so I bought a college algebra book. The book is "McGraw Hill - College Algebra Demystified" and it seems like a pretty good book, but lacks a lot explanations. I have done quadratic equations before, but not radicals with fractions.

This is the problem: 3x^2+9x-2=0

I'm on the first chapter of this book and I'm completing the square and I don't have any problem there. The problem is with my answer and their extra step with the fraction.

I end up with: x = -3/2 + √35/12 and -3/2 - √35/12

The books answer is: x = -3/2 + √105/6 and -3/2 - √105/6

I understand that you get: 2√3 for 12. But why are the numerator and denominator both multiplied by 3?

I feel crazy asking this question, because I can see how it's done, but I need an explanation on the last part.
Please, please, please use parentheses to clarify what you are saying! In the "your" answer you have √(35/12), NOT √35/12= (√35)/12! √(35/12)= √(35)/√(12)= √(35)/(2√3). Now, they did NOT multiply numerator and denominator by 3, they multiplied by √3 in order to "rationalize" the denominator: (√3)(√(35))/(√3)(2√3)= √(105)/6.

8. Originally Posted by HallsofIvy
Please, please, please use parentheses to clarify what you are saying! In the "your" answer you have √(35/12), NOT √35/12= (√35)/12! √(35/12)= √(35)/√(12)= √(35)/(2√3). Now, they did NOT multiply numerator and denominator by 3, they multiplied by √3 in order to "rationalize" the denominator: (√3)(√(35))/(√3)(2√3)= √(105)/6.
If you read his question, you would see that they in fact did multiply by three.

'But why are the numerator and denominator both multiplied by 3?'

He didn't say anything about root 3.

They did it this way because the cancellation (by division, not multiplication unless your talking about reciprocols) was implied.