What are the solutions of the equation 48/(2x+2)=2x+6?

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neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

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To find the solutions of 48/(2x+2) = 2x+6.

We  multiply both sides by 2x+2 .

48 = (2x+2)(2x+6)

48 = 2(x+1)(2)(x+3)

48 = 4  (x+1)(x+3)

We divide  both sides to keep the equation lower terms.

12 = (x+1)(x+3)

We expand the right side:

12 = x^2 +4x+3.

Subtracr 12:

0 = x^2 +4x +3-12.

x^2 +4x-9 = 0

 x1 = {-4 + sqrt(4^2+4*9)}/2 = {-4 +sqrt 52}/2 = {-4+2sqrt(13)}/2 = (-2+sqrt13).

x 1 = (-2+sqrt13)

x2 = (-2-sqrt13).

Therefore the solutions of the given equation are (-2+sqrt13) and  -(2+ sqrt13)

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll factorize both sides by 2:

2*24/2(x+1) = 2(x+3)

We'll simplify:

24/(x+1) = 2(x+3)

We have to isolate x to the left side. For this reason, we'll have to multiply both sides by the denominator x+1.

24(x + 1)/ (x + 1) = 2(x+3)(x+1)

We'll simplify and we'll get:

2(x+3)(x + 1) = 24

We'll divide by 2 both sides:

(x+3)(x + 1) = 12

We'll remove the brackets:

x^2 + x + 3x + 3 - 12 = 0

We'll combine like terms:

x^2 + 4x - 9 = 0

The number of the roots is 2 and the formula for finding the roots is:

x1 = [-b+sqrt(b^2 - 4ac)]/2a

x2 = [-b-sqrt(b^2 - 4ac)]/2a

Let's identify a,b,c:

a = 1

b = 4

c = -9

x1 = [-4+sqrt(16+36)]/2

x1 = (-4+sqrt52)/2

x1 = 2(-2+sqrt13)/2

x1 = -2+sqrt13

x2 = -2-sqrt13

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