solving equations by factorising...please HELP!?

Question

How do i solve equations by factorising in the form of ax2 (that 2 is supposed to be [squared]) + bx+c=0

can you take me thru step by step of how to solve this: 3x2 + 10x + 3 = 0
(The 2 after 3x is supposed to be a little 2 for squared)

I dont understand anything u guys are saying! Im 15...can u please put it simpler for me :)

Answer 1

you have to give your eqation to solve.

Q) 3x2 + 10x + 3 = 0

A) the above can be written as

step 1) 3x2 + 9x +x + 3 = 0 (as 9x+x=10x)

step2) (3x2 + 9x) + (x+3)=0 (separated 1st two and last
two with brakets)

step 3) 3x(x +3) + (x+3) = 0 (take 3x common from 1st factor

step 4) (x+3)(3x+1) = 0 (take x+3 common)


step 5) (x+3)=0 or (3x+1)=0

step 6) x=-3 or x=-1/3

That is the answer

email me if it is still not clear.

:-)

Answer 2

Here are certain aids to help you solve quadratic equations using factorization:

1. If all terms are positive terms - ie ax² + bx + c, the 2 factors have the form (a1x + c1) (a2x + c2)

2. If the middle term is negative and the third term positive - ie ax² - bx + c, the 2 factors have the form (a1x - c1)(a2x - c2)

3. And lastly, both 2nd and 3rd terms are negative, ie ax² - bx - c, the 2 factors have the from (a1x + c1)(a2x - c2) or (a1x - c1) (a2x + c2)

4. The rest is brute force. If a = 1, then just figure out which factors of c works such that c1 +/- c2 = b.

5. It is a little more complicated when a > 1. Factorize a and c such that a1*c2 +/- a2*c1 = b

Good luck!

Other methods for solving quadratics:

1. Completing the square
2. Difference of 2 squares
3. Using the quadratic formula

Answer 3

your answer will be in the form of (x+n)(x+m) where n and m are just numbers.

this takes practice to see what numbers are the ones to pick-- xn + xm have to equal the bx term, and n*m need to equal the c term. So what i do is look for what two numbers can be multiplied to get "c", then find a way to add or subract those two numbers to get the "b" term.

ex. x^2+5x+4

4 = 2*2, 4*1-- pick the 4*1 b/c 4 + 1 = 5 (which is the b term)

then you get (x+1)(x+4) = 0

then its easy-- you set both parts in parentheses equal to 0
x+1=0, x+4=0

you get that the answers are -1, and -4

Answer 4

an equation of this type is called quadratic equation..
to solve the equations of quadratic equations,we have to follow the following steps:-
In ax^2(here 2 is supposed to be square)
SO,in ax^2+bx+c,we have to find two numbers of which product is equal to 'c' and their sum is equal to 'b'
For eg,Q1) x^2+5x+6
Ans: (x+2)(x+3) {here 2*3=6 and 2+3=5}

Q2)2x^2+x-6
Ans) the first step here is to mutiply 2(the coefficient of x^2) with the constant {that's mutilpying a with c in your equation)
so now we have two numbers such that their product is -12 and their sum is 1
the numbers therefore will be 4 and -3..
hence,2x^2+4x-3x-6
Taking the common factors outside, 2x(x+2)-3(x+2)
Therefore the final answer (2x-3)(x+2)


So in any quadratic equations,above are the steps for factorising
Hope i have helped you...
Enjoy solving......... :)

Answer 5

ax^2+bx+c=0
multiply both sides by 4a
4a^2x^2+4abx+4ac=0
4a^2x^2+4abx=-4ac
completing the squareon he LHS
(2ax)^2+2*2ax*b+(b)^2=-4ac+b^2
(2ax+b)^2=b^2-4ac
2ax+b=+/-sq.rt.(b^2-4ac)
2ax=-b+/-sq.rt.(b^2-4ac)
x=[-b+/-sq.rt.(b^2-4ac)]/2a
so the factors are
{x-{-b+sq.rt(b^2-4ac)]/2a}{x-[-b-sq.rt(b^2-4ac)]/2a}=0

Answer 6

ax^2+bx+c=0

x= (-b+-Sqr(b^2-4ac))/2a

Answer 7

So ax^2+bx+c = a*(x-x1)*(x-x2), where
x1={-b-sqrt(b^2-4ac)}/(2a),
x2={-b+sqrt(b^2-4ac)}/(2a)