1. ## algebra

Can anybody please help me out on how to solve this type of problem? The direction are Solve. Find the sum of the two roots. x^+3x+2=0. Thanks!

2. Originally Posted by fulloflove35
Can anybody please help me out on how to solve this type of problem? The direction are Solve. Find the sum of the two roots. x^+3x+2=0. Thanks!
I suppose you meant x^2 + 3x + 2 = 0

we can factorize this guy. think of two numbers that when multiplied yields +2 and when added yields +3. +1 and +2 works, so:

x^2 + 3x + 2 = 0

=> (x + 2)(x + 1) = 0

=> x + 2 = 0 or x + 1 = 0 .........since if the product of two numbers give zero, one or the other number has to be zero

=> x = -2 or x = -1

these are the two roots. and the sum is .......?

alternate method:

for any quadratic y = ax^2 + bx + c

the sum of the roots is: -b/a

the product of the roots is: c/a

3. ## thank you

Thanks for helping can you please help me with a few more? I would greatly appreciate it. I did not get the first step at all. Factorizing?

1) X^2-x-6=0, 2) x^2+5x+4=0, 3) x^2-5x+6=0 What do you do when the signs are different? like in problem number three.

4. Originally Posted by fulloflove35
Thanks for helping can you please help me with a few more? I would greatly appreciate it. I did not get the first step at all. Factorizing?

1) X^2-x-6=0, 2) x^2+5x+4=0, 3) x^2-5x+6=0 What do you do when the signs are different? like in problem number three.

When you factor a quadratic whose leading coefficient is 1 (which is true in all these cases), simply follow these steps: I'll use problem #3 as my example.

1. Determine what two factors when multiplied together = $\displaystyle 6$ and when added together =$\displaystyle -5$. That's it!

Factors $\displaystyle 3\times2= 6$, and sum $\displaystyle 3+2=5$==>No Good! We need $\displaystyle -5$

Factors $\displaystyle 6\times1=6$, and sum $\displaystyle 6+1=7$==>No Good! We still need $\displaystyle -5$

Factors $\displaystyle -3\times-2=6$, and sum $\displaystyle -3+(-2)=-5$==>Good, we found them.

Therefore,

$\displaystyle x^2-5x+6=0$

$\displaystyle (x-3)(x-2)=0$

Use the zero product property to obtain:

$\displaystyle x-3=0 \ or \ x-2=0$

Finally,

$\displaystyle x=3 \ or x=2$

Try this technique on the other problems you have. When you encounter quadratics whose leading coefficients are not 1, then we'll talk about another technique. First things first.