Quadratic Equations?

1)
4x^2 + 3x - 1
= 4x^2 + 4x - x - 1
= 4x(x + 1) - 1(x + 1)
= (x + 1)(4x - 1)

2)
3x^2 + 2
= cannot be factored

3)
3x^2 + 5x - 2
= 3x^2 + 6x - x - 2
= 3x(x + 2) - 1(x + 2)
= (x + 2)(3x - 1)

= = = = = = = =

4)
x^2 - 6x + 9 = 16
x^2 - 6x + 9 - 16 = 0
x^2 - 6x - 7 = 0
x^2 + x - 7x - 7 = 0
x(x + 1) - 7(x + 1) = 0
(x + 1)(x - 7) = 0

x + 1 = 0
x = -1

x - 7 = 0
x = 7

∴ x = -1 , 7

= = = = = = = =

5)
(x + 3)^2 - 3 = 13
(x + 3)^2 = 13 + 3
(x + 3)^2 = 16
x + 3 = ±√16
x + 3 = ±4

x + 3 = 4
x = 4 - 3
x = 1

x + 3 = -4
x = -4 - 3
x = -7

∴ x = -7 , 1

I'm assuming that when you refer to "a" you're using the convention:

ax^2 + bx + c

If so, the first thing to look at is the a.

1. 4x^2 + 3x - 1

In this case a = 4. So, we find pairs of numbers that multiply together to give 4. 1 and 4, or 2 and 2 are all we get. You'll find quickly that 2 and 2 won't work. So, let's use 4 and 1.

Write this down:

(4x )(x )

Now we need to see what else goes in the parenthesis. To do so we'll look at c.

We really only have one option. 1 and -1 are the only pairs that multiply together to get -1. Put the 1s in now:

(4x 1)(x 1)

Now we just have to figure out which one is minus and which one is plus. We get that by looking at b. The product of the 4x and its 1 plus the product of the 1x and its 1 have to add to 3. Basically we'll get a 4x out and a 1x out, and we get to pick which one is positive and which one is negative. So, we have to pick the way that gives us the positive 3x. Clearly, the 4x has to be multiplied by the positive 1, which means the negative 1 must multiply with the x. And hence the answer is:

(4x - 1)(x +1)

2. 3x^2 + 2

This one doesn't factor at all.

3. 3x^2 + 5x - 2

Using the same convention as in #1. Look at a, only 3 and 1 work:

(3x )(x )

Now look at c. We need a 2 and a 1, but one of them is negative and one is positive. So, before we put them in we have to figure out which spot the go in as well as their signs.

So, look at b. to get a positive 5 we're going to have to multiply the 2 by the 3x, and the 2 must be positive. Which makes the 1 negative and together they'll give us the 5 we need:

(3x - 1)(x + 2)

4. x^2 - 6x + 9 = 16

First step here is to subtract 16 from both sides:

x^2 - 6x - 7 = 0

Now we just follow the convention as before a=1, so 1 and 1 are our only choice:

(x )(x )

Look at c, 7 and 1 are our only choice. We don't need to figure out which one goes with which x since we just have 1x in each set of parenthesis. Just toss them in and we'll figure the signs later:

(x 7)(x 1)

b = -6, so we'll have better luck adding +1 to -7 than vice versa:

(x - 7)(x + 1)

5. (x+3)^2 - 3 = 13

Here we have to expand the (x+3)^2 and subtract 13 from both sides:

(x+3)(x+3) - 3 - 13 = 0

x^2 + 6x - 7

From here we have the same problem as #4 except the 6x is positive. So the problem is the same up to picking the signs:

(x 7)(x 1)

This time the 7 has to be positive and the 1 negative to give the +6x we need:

(x + 7)(1 - x)

Have fun.

1) (4x-1)(x+1)

2) cannot factor

3) (3x-1)(x+2)

4)x^2-6x-7=0
(x-7)(x+1)

5) x^2+6x+9-3=13
x^2+6x+6=13
x^2+6x-7=0
(x+7)(x-1)

to solve these quadratic eqn follow these steps i'll do it with example
take ur first eqn 4x^2+3x-1;
first multiply coefficients of x^2 and the constant i.e 4*(-1)= - 4
now split the coefficients of x i.e 3 in such a way that its product is -4 and sum is 3
its 4,-1
so
4x^2+4x-x-1 =>4x(x+1)-(x+1) =>(x+1)(4x-1)
so the factors are x+1 and 4x-1

there is another simple formula
for eqn ax^2+bx+c=0;
x= (-b+(or -)d) / (2*a);
where d=sqrt(b^2 - 4*a*c)

---------
3)(3x-1)(x+2)

1. 4x^2 + 3x - 1
a=4 b=3 c=-1

2. 3x^2 + 2
a=3 b=0 c=2

3. 3x^2 +5x - 2
a=3 b=5 c=-2

4. x^2 - 6x + 9 = 16
x^2 - 6x + 9 -16=0
x^2 - 6x -7 = 0
a=1 b=-6 c=-7

5.(x+3)^2 - 3 = 13
x^2+6x+9-3-13=0
x^2+6x-7 =0
a=1 b=6 c=-7

Hope it helps. W.

1. 4x^2 + 3x - 1
4x^2+4x-x-1
4x(x+1)-1(x+1)
(x+1)(4x-1)=0
x=-1, 4x-1 =0 gives x=1/4

3. 3x^2 +5x - 2
To factor this, try to find 2 numbers such that their sum is 5 and their product is -6.
3x^2+6x-x-2
3x(x+2)-1(x+2)
(x+2)(3x-1)
If you solve this (x+2)(3x-1)=0
x=-2 and x=1/3

First state the unknown is a or x. Edit the question. Or is a=x?
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