how do i solve the following equations:(factoring) 1. 2x^2+3x-2=0 2. 2x^2-3x-9=0 3. 4x^2-8x+3=0 4. 9x^2-4=0 i need to know how to write out...

1. 2x^2+3x-2=0

We do the
factorisation by groupig method by splitting the middle term  3x= -4x+x and then equate each
factor to zero and solve for x.

 2x^2-4x+x-2 =
0.

2x(x-2)+1(x-2) = 0.

(x-2)(x+1) =
0.

Therefore x-2 = 0 or x+1 = 0.

x= 2,
or x=-1.

2. 2x^2-3x-9=0.

We do this by
completing the square .

2x^2-3x-9 =
0

Divide by
2:

x^2-3x/2+9/2

(x-3x/2
+(3/4)^2)-(3/4)^2-9/2 = 0

(x-3/4)^2 = 9/16+72/16 =
81/16

x-3/4 = 9/4 , or x-3/4 =
-9/4.

 x= 3/4+9/4 = 3, or x= 3/4-9/4= -6/4 =
-1.5.

3. 4x^2-8x+3=0.

We complete the
square 4x^2-8x.

4x^2-8x+4-4 +3 =
0

(2x-2)^2-1 = 0

(2x-2)^2 =
1

2x-2 = 1, or 2x-2= -1.

2x=2+1=3 ,
or 2x= 2-1 = 1.

x= 3/2 or x= 1/2.

4.
9x^2-4=0.

Add 4 to both sides:

9x^2=
4.

x^2= 4/9.

x = sqrt(4/9), or x= -
sqrt(4/9).

x= 2/3 or x= -2/3.