How do u solve x (squared)+3x=4?

Question

please explain

Answer 1

Well it's easiest for the sake of factoring to make your equation equal 0.

So you want to subtract 4 from both sides.  That makes it look like this: x^2 + 3x - 4 = 0

Then you need to factor.  So you can make it into the product of 2 quantities.

(x +/- ?)(x +/- ?) = 0

Well you need the product of those second numbers in each quantity to equal -4 and have a sum of 3.  -1 and 4 should do the trick.  This turns the equation into this: (x - 1)(x + 4) = 0

Now, for it to equal 0, x would have to be the opposite of one of the numbers in the quantities.  So that gives 1 or -4, which would then be your solution set.

x = {1, -4}

Hope that helps!

Answer 2

X^2  + 3X -4 = 0,(X+4)(X-1)=0,then eighter one could be zero,

X+4=0 X=-4   &    X-1=0   X=1CHECK: 1(squared) + 3(1)=4

and (-4) squared + 3(-4) =4

Answer 3

X^2+3X =  4

Subtract 4 from each side

X^2 + 3X - 4 = 0

Factor the left hand side:

(X+4)(X-1) = 0

This equation is true whenever X = -4 or X =1.

You solution set is { -4, 1 }

A previous answerer factored the left hand expression incorrectly.

CHECK:

(-4)^2 + 3(-4)  =  16 -12  =  4

(1)^2  + 3(1)    =   1 +  3  = 4

Answer 4

x^2 + 3x - 4 = 0

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

x = 4, -1

Hope tht helps :)

Answer 5

x² + 3x - 4 = 0

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

x = - 4 , x = 1