D dayton1 New member Joined Oct 22, 2007 Messages 33 Oct 30, 2007 #1 I have to solve for x for this equation......but I can't seem to do it right! 2x^2+5x+6= x^2+5x+6 It seems that no matter what you do you eliminate the x and end up with only the x^2 numbers.....any help?
I have to solve for x for this equation......but I can't seem to do it right! 2x^2+5x+6= x^2+5x+6 It seems that no matter what you do you eliminate the x and end up with only the x^2 numbers.....any help?
O o_O Full Member Joined Oct 20, 2007 Messages 393 Oct 30, 2007 #2 Don't worry. You're on the right track. Keep going with what you have: 2x² = x² 2x² - x² = 0 ... Can you see the end result is?
Don't worry. You're on the right track. Keep going with what you have: 2x² = x² 2x² - x² = 0 ... Can you see the end result is?
D dayton1 New member Joined Oct 22, 2007 Messages 33 Oct 30, 2007 #3 ok so then I have x(2x-x)=0 , Right???? Then what! arrghh :x .
J jwpaine Full Member Joined Mar 10, 2007 Messages 719 Oct 30, 2007 #4 \(\displaystyle \L 2x^2+5x+6= x^2+5x+6\) Subtract all right hand side terms from both sides \(\displaystyle \L 2x^2-x^2+5x-5x+6-6 = 0\) \(\displaystyle \L x^2 = 0\)
\(\displaystyle \L 2x^2+5x+6= x^2+5x+6\) Subtract all right hand side terms from both sides \(\displaystyle \L 2x^2-x^2+5x-5x+6-6 = 0\) \(\displaystyle \L x^2 = 0\)