How do you solve a quadratic equation where b = the sum of square roots?

**gavinhudson** · October 15th 2010, 02:03 AM

I understand the quadratic formula. I can derive the quadratic formula. I do not understand what to do when b= the sum of two square roots, as in the following problem:

Use the quadratic formula to solve:

r squared + [square root (3) - square root (2)]*r = square root (6)

r^2+(√(3)-√(2))r = √(6)

I know the answer is r = -(√(3)) or √(2). I don't know how to get to that answer. I have tried a number of ways. Plugging a, b, and c straight into the quadratic formula, I get:

r = 1/2[√(2)-√(3)+√(5+(2√(6)))] or 1/2[√(2)-√(3)-√(5+(2√(6)))]

This is right, but I don't understand the following steps:

1/2[√(2)-√(3)+√(5+(2√(6)))] --> √(2)
1/2[√(2)-√(3)-√(5+(2√(6)))] --> -(√(3))

Thanks for your help! (I've been working on this problem for WAY too long.)

**Prove It** · October 15th 2010, 02:23 AM

$r^2 + (\sqrt{3} + \sqrt{2})r = \sqrt{6}$

$r^2 + (\sqrt{3} + \sqrt{2})r - \sqrt{3}\sqrt{2} = 0$

$\displaystyle{r = \frac{-(\sqrt{3} + \sqrt{2}) \pm \sqrt{(\sqrt{3} + \sqrt{2})^2 - 4(1)(\sqrt{3}\sqrt{2})}}{2(1)}}$

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

$\displaystyle{= \frac{-\sqrt{3} - \sqrt{2} \pm \sqrt{(\sqrt{3})^2 - 2\sqrt{3}\sqrt{2} + (\sqrt{2})^2}}{2}}$

$\displaystyle{= \frac{-\sqrt{3}-\sqrt{2} \pm \sqrt{(\sqrt{3} - \sqrt{2})^2}}{2}}$

$\displaystyle{= \frac{-\sqrt{3} - \sqrt{2} \pm (\sqrt{3} - \sqrt{2})}{2}}$

$\displaystyle{r = \frac{-2\sqrt{2}}{2}}$ or $\displaystyle{r = \frac{-2\sqrt{3}}{2}}$

$\displaystyle{r = -\sqrt{2}}$ or $\displaystyle{r = -\sqrt{3}}$ .

**gavinhudson** · October 15th 2010, 05:46 PM

This answer helped me solve the problem! Thank you. The question answered here is a little different from mine. This question answers r^2+(√(3)+√(2))r = √(6) instead of r^2+(√(3)-√(2))r = √(6), but it was really helpful as an example format to use to solve the latter question. The key that I was missing was factoring √((√3)^2 + 2√(3)√(2) + (√(2))^2) ---> √((√(3) + √(2))^2). Thank you again!

**Wilmer** · October 15th 2010, 06:32 PM

Originally Posted by Prove It

$r^2 + (\sqrt{3} + \sqrt{2})r = \sqrt{6}$
$\displaystyle{r = -\sqrt{2}}$ or $\displaystyle{r = -\sqrt{3}}$ .

Should be [SQRT(3) - SQRT(2)]r
So answer is r = SQRT(2) or r = -SQRT(3)

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