Question

solve by completing the sq

3x^2-12x+11=0

Answer 1

do u know how to solve this problem......
u need to know the procedure

Answer 2

divide by 3 and take the constant to the other side, then add (b/2)^2 to both sides
THat is the procedure

Answer 3

yes i need step by step please

Answer 4

\[3x^2-12x=-11\]
\[x^2-4x=-\frac{11}{4}\]
\[(x-2)^2=-\frac{11}{2}+4=-\frac{3}{2}\]
is it ok if the solutions are complex?

Answer 5

that is wrong satellite

Answer 6

step one
subtract 11 from both sides
step 2, divide by 3
yes is certainly made a mistake there!

Answer 7

no trouble there he is bang on!

Answer 8

should be
\[x^2-4x=-\frac{11}{3}\]

Answer 9

in second step he is adding 4 on both sides

Answer 10

then take half the coefficient of the x term (here -4) square it and add to the right to give
\[(x-2)^2=-\frac{11}{3}+4=\frac{1}{3}\]

Answer 11

then take the square root of both sides. you will get two solutions
\[x-2=\sqrt{\frac{1}{3}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\] and
\[x-2=-\frac{\sqrt{3}}{3}\]

Answer 12

finally add 2 to both sides to get your answers. they are
\[x=2-\frac{\sqrt{3}}{3}\] and
\[x=2+\frac{\sqrt{3}}{3}\]

Answer 13

thank you you are very helpful:)