How do you simplify #x^2/( x+12) = (2/1)#?

Question

Nathan Axel · Accepted Answer

Assuming you mean to solve the equation, the solutions are #x=-4# and #x=6#.

You mean solve? To solve the equation, first of all note that #2/1=2#, and that we will not accept #x=-12# as a solution, because it would annihilate the denominator. So, if #x# is not #-12#, we can multiply the whole expression by #x+12#, obtaining #x^2=2(x+12)#. Expanding the right member we get #x^2=2x+24#, and bringing everything to the left member we have #x^2-2x-24=0#, which is a classical quadratic equation of the form #ax^2+bx+c=0#, where #a=1#, #b=-2# and #c=-24#. The discriminant of the quadratic is #b^2-4ac#, in your case 4-4*(-24)= 4+96=100#, which is positive, so the equation has two solutions, namely #x_{1,2} = \frac{-b\pm\sqrt(b^2-4ac)}{2a}= (2pmsqrt(100))/2=(2pm10)/2=(1pm5)# So, the two solutions are #-4# and #6#.

Ethan Adkins · Answer

undefined To simplify the equation x^2/(x+12) = 2/1, we can cross-multiply to eliminate the fractions. This gives us x^2 = 2(x+12). Expanding the right side, we have x^2 = 2x + 24. Rearranging the equation, we get x^2 - 2x - 24 = 0. Factoring the quadratic equation, we have (x-6)(x+4) = 0. Setting each factor equal to zero, we find x = 6 or x = -4. Therefore, the solutions to the equation are x = 6 and x = -4.