# EquationProvide 2 methods to solve the equation x^2 + 3x – 4 = 0?

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Another method to solve x^2 + 3x – 4 = 0 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 = 1 and x = -4

**The solution of the equation is x = 1 and x = -4**

One method is to complete the square

x^2 + 3x + _

We'll use the formula for a binomial raised to square:

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

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

b^2 = 9/4

To get a perfect 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'll get 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 requested solutions of the equation are {-4 ; 1}.