1. ## absolute values and quad. formulae

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

Therefore, $a = \frac{(m+3) + \sqrt{(m+3)^2 - 4(2-m)}}{2}$ and $b = \frac{(m+3) - \sqrt{(m+3)^2 - 4(2-m)}}{2}$, since we let $a > b$

$|a|=|b+1|$

$\frac{(m+3) + \sqrt{(m+3)^2 - 4(2-m)}}{2} = \left|\frac{(m+3) - \sqrt{(m+3)^2 - 4(2-m)}}{2}\right| + 1$

You can now solve for m.

3. Originally Posted by colby2152
since $a > b$
So, as $|-2|=|1+1|,\,-2>1$ ?

4. Originally Posted by flyingsquirrel
So, as $|-2|=|1+1|,\,-2>1$ ?
Okay, we let $a > b$

5. Originally Posted by colby2152

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

Therefore, $a = \frac{(m+3) + \sqrt{(m+3)^2 - 4(2-m)}}{2}$ and $b = \frac{(m+3) - \sqrt{(m+3)^2 - 4(2-m)}}{2}$, since we let $a > b$

$|a|=|b+1|$

$\frac{(m+3) + \sqrt{(m+3)^2 - 4(2-m)}}{2} = \left|\frac{(m+3) - \sqrt{(m+3)^2 - 4(2-m)}}{2}\right| + 1$

You can now solve for m.
i still don't understand. Why
$|a|=|b+1|$
$\frac{(m+3) + \sqrt{(m+3)^2 - 4(2-m)}}{2} = \left|\frac{(m+3) - \sqrt{(m+3)^2 - 4(2-m)}}{2}\right| + 1$
shouldn't squaring the both sides is the only way to do it?
$a^2=(b+1)^2$
thanks for help!

6. Originally Posted by afeasfaerw23231233
i still don't understand. Why
$|a|=|b+1|$
$\frac{(m+3) + \sqrt{(m+3)^2 - 4(2-m)}}{2} = \left|\frac{(m+3) - \sqrt{(m+3)^2 - 4(2-m)}}{2}\right| + 1$
shouldn't squaring the both sides is the only way to do it?
$a^2=(b+1)^2$
thanks for help!
No, keep the absolute value sign on the right side because there could be a negative.

7. Originally Posted by afeasfaerw23231233
i still don't understand. Why
$|a|=|b+1|$
$\frac{(m+3) + \sqrt{(m+3)^2 - 4(2-m)}}{2} = \left|\frac{(m+3) - \sqrt{(m+3)^2 - 4(2-m)}}{2}\right| + 1$
shouldn't squaring the both sides is the only way to do it?
$a^2=(b+1)^2$
thanks for help!
Denoting it by

$x + y = \left|x - y \right| + 1$
Consider two cases x > y and x <= y and then reduce it to $y = \frac12$ and $x = \frac12$ respectively.

Now consider each case separately, and solve for m and check if the assumed condition (x > y or x <= y) satisfies.