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

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We have to solve the quadratic equation x^2 + 3x – 4 = 0.

One way to do this is by factorization.

x^2 + 3x – 4 = 0

=> x^2 + 4x – x – 4 =0

=> x(x+4) -1(x+4) = 0

=> (x – 1)(x + 4) =0

=> x can be 1 and -4

Another way to solve this is to use the formula for roots of a quadratic equation which is [–b+ sqrt(b^2 – 4ac)] / 2a and –b - sqrt(b^2 – 4ac)] / 2a

substituting the values we get

x1 = [-3 + sqrt(9 + 16)]/ 2

=>( -3 + 5)/2 = 2/2 = 1

x2 = [-3 - sqrt(9 + 16)]/ 2

=>( -3 - 5)/2 = -8/2 = -4

**The roots of the equation are x = 1 and x = -4.**

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.

Another way is to complete the square

x^2 + 3x + _

The formula of the binomail raised to square is:

(a+b)^2 = a^2 + 2ab + b^2

a = x

2xb = 3x => b = 3/2

b^2 = 9/4

To complete the square, we'll add and subtract 9/4:

(x^2 + 3x + 9/4) - 9/4 – 4 = 0

(x + 3/2)^2 - 25/4 = 0

We notice that is a difference of squares:

a^2 - b^2 = (a-b)(a+b)

We'll put a = x + 3/2

b = 5/2

(x + 3/2)^2 - 25/4 = (x + 3/2 - 5/2)(x + 3/2 + 5/2)

(x + 3/2 - 5/2)(x + 3/2 + 5/2) = 0

We'll put each factor as zero:

x + 3/2 - 5/2 = 0

We'll combine like terms:

x - 2/2 = 0

x - 1 = 0

x = 1

x + 3/2 + 5/2 = 0

x + 8/2 = 0

x + 4 = 0

x = -4

**The solutions of the equation are {-4 ; 1}.**