find the roots of (x-2)/(x+2)=(1)/(x-3)?

Question

stuck on a math problem, i dont know where to start x.x

find the roots (x-2)/(x+2)=(1)/(x-3)

Answer 1

... (x-2) / (x+2) = (1) / (x-3) ← x ≠ -2 or 3 else entire equation becomes undefined

or (x-2) (x-3) = (1)(x+2)

or x^2 - 5x + 6 = x + 2

or x^2 - 6x + 4 = 0

or x^2 - 6x + 4 = 0

Use the quadratic formula:

x = [ -b ± √ ( (b)² - 4(a)(c) ) ] / 2(a)

x = [ -(-6) ± √ ( (-6)² - 4(1)(4) ) ] / 2(1)

x = [ 6 ± √ ( (36 - 16) ] / 2

x = [ 6 ± √20 ] / 2

x = [ 6 ± 2√5 ] / 2

x = 3 ± √5

x = {3-√5, 3+√5) ~ (0.7639, 5.2361)

Answer 2

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



Cross multiply:

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

x^2 - 5x + 6 = x + 2

x^2 - 6x = -4

x^2 - 6x + 9 = 9 - 4

(x - 3)^2 = 5

x - 3 = ±√5



x = 3 ± √5

Answer 3

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

=> (x-2)(x-3) = (x+2)

=> x² -5x +6 = x + 2

=> x² -6x + 4 = 0

=> x1 =√5 + 3 , x2 = √5 - 3

Answer 4

Multiply both sides by (x+2)(x-3). That will give you a polynomial, which you should be able to solve.

Answer 5

x=3-sqrt(5)

x=3+sqrt(5)

Answer 6

(x-3)(x-2)(x+2)-1=0;

(x^2-4)(x-3)-1=0;

x^3-3x^2-4x+11=0;

x(x^2-3x-4)+11=0;

y=x-1;

y^3+px+q=0;

p=3-4=-1;

q=2(-3)^3/27-3*4/3+11=5;

Q=-1/3+25/4;

A=sqrt(3)(-5/2+sqrt(Q));

B=sqrt(3)(-5/2-sqrt(Q));

y1=A+B; y2,3=-(A+B)/2+-i(A-B)sqrt(3)/2;

i=sqrt(-1);

x1,2,3=y1,2,3+1