Question

solving quadratic equations by taking square roots (4x+1)^2-8=0 why do you add 8 to both sides? is it because you cant take the square root of a negative? what if the equation said (4x+1)^2=-8? what would you do then?

Answer 1

try an equation with actual numbers - that might help you understand what is going on.
e.g.\[15-8=7\]so if we add 8 to both sides we get:\[15-8+8=7+8\]\[15=15\]in your quadratic:\[(4x+1)^2-8=0\]in order to find x, we need to get all terms NOT involving x to the right-hand-side of this equation, so we add 8 to both sides to get:\[(4x+1)^2-8+8=0+8\]\[(4x+1)^2=8\]then take square roots of both sides to get:\[4x+1=\pm\sqrt{8}=\pm\sqrt{4*2}=\pm2\sqrt{2}\]so:\[4x=-1\pm2\sqrt{2}\]\[x=\frac{-1\pm2\sqrt{2}}{4}\]

on your other question, if you had something like:\[x^2+16=0\]then you would subtract 16 from both sides to get:\[x^+16-16=0-16\]\[x^2=-16\]then you can still take square roots of both sides to get:\[x=\pm\sqrt{-16}=\pm4i\]where i is the square root of -1 (called an imaginary number).

Answer 2

oh i see... so you only added 8 to the other side to get the x alone??

Answer 3

yes - that is right - I hope you understand it better now.

Answer 4

thanks and yes i do !

Answer 5

yw