Solve the equation x^2 + 3x – 4 = 0 in two ways.

To solve  x^2+3x-4 =0.

Add 4 to
both sides:

x^2+3x= 4.

Add (3/2)^2 to
both sides:

x^2+3x+(3/2)^2 = 4+(3/2) =
25/4.

(x+3/2)^2 = (5/2)^2.

We take
sqare root on both sides:

So x+3/2 = 5/2 , or x-5/2 =
-5/2.

x =  5/2-3/2 = 1 . Or x = -5/2-3/2 = -8/2 =
-4.

Second method:

x^2+3x-4 =
0

Let  Let x1+x2 be roots.

By the
relation between the roots and coefficients, we have:

Then x1+x2 =
-(3)/1 and x1x2 = -4, where x1 and x2 are the roots of x^2-3x-4 =
0.

Therefore x1+x2 = -3 and x1x2 =
-4.

Therefore (x1 - x2) = sqrt {(x1+x2)^2 - 4x1x2 } = sqrt{(-3)^2 -
4(-4)} = sqrt(9+16) = sqrt25 = 5.

Therefore x1+x2 = -3...(1) , 
(x1-x2) = 5.....(2).

(1)+(2) gives: 2x1 = -3+5 = 2. So x1 = 2/2 =
1.

(1)-(2) gives: 2x2 = -3-5 = -8. So x2 = -8/2 =
-4.

Therefore in both cases the roots of the given equation are  1
and -4.