S sarahj3 New member Joined Apr 4, 2006 Messages 10 May 22, 2006 #1 Solve: 1/8x^2+1/5x=-30/40 40*1/5x^2=3*40/40 5x^2+8x=-3 -8+or- the square root of 64-4*5*3/2*5 -8+or-8 the square root of 4/10 -8+or- 2/10=-6/10 x1=-3/5 x2=-10/10 =-1
Solve: 1/8x^2+1/5x=-30/40 40*1/5x^2=3*40/40 5x^2+8x=-3 -8+or- the square root of 64-4*5*3/2*5 -8+or-8 the square root of 4/10 -8+or- 2/10=-6/10 x1=-3/5 x2=-10/10 =-1
J jsbeckton Junior Member Joined Oct 24, 2005 Messages 174 May 22, 2006 #2 if you have x^2=-10/10 which means x=the square root of a negative number. have you talked agout "i", imaginary numbers? and to "plug in" to check your anwser, just put a -1 everywhere you have an x
if you have x^2=-10/10 which means x=the square root of a negative number. have you talked agout "i", imaginary numbers? and to "plug in" to check your anwser, just put a -1 everywhere you have an x
J jsbeckton Junior Member Joined Oct 24, 2005 Messages 174 May 22, 2006 #3 sarahj3 said: Solve: 1/8x^2+1/5x=-30/40 40*1/5x^2=3*40/40 Click to expand... Why are you multiplying 1/5x^2 by 40? You don't even have a 1/5x^2 in your equation 1/8x^2+1/5x=-30/40 Click to expand... If you want to reduce you have to multiply EACH term by 40.
sarahj3 said: Solve: 1/8x^2+1/5x=-30/40 40*1/5x^2=3*40/40 Click to expand... Why are you multiplying 1/5x^2 by 40? You don't even have a 1/5x^2 in your equation 1/8x^2+1/5x=-30/40 Click to expand... If you want to reduce you have to multiply EACH term by 40.
