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 = $6$ and when added together = $-5$. That's it!

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

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

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

Therefore,

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

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

Use the zero product property to obtain:

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

Finally,

$x=3 \ or x=2$

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.

Hope this has been helpful.