Question

completing the square.- x^2+3x-18=0 so..x^2+3x_=18+_..... _ = ?

Answer 1

why -6,3 ?!

Answer 2

Is the expression \[\large - x^2+3x-18=0\] or is it \[\large x^2+3x-18=0\] ???

Answer 3

2nd one

Answer 4

\[\large x^2+3x-18=0\]


\[\large x^2+3x=18\]


\[\large x^2+3x+\frac{9}{4}=18+\frac{9}{4}\]


\[\large (x+\frac{3}{2})^2=\frac{81}{4}\]


\[\large x+\frac{3}{2}=\pm\sqrt{\frac{81}{4}}\]


\[\large x+\frac{3}{2}=\pm\frac{9}{2}\]


\[\large x=-\frac{3}{2}\pm\frac{9}{2}\]


\[\large x=-\frac{3}{2}+\frac{9}{2} \ \text{or} \ \ x=-\frac{3}{2}-\frac{9}{2}\]


\[\large x=\frac{6}{2} \ \text{or} \ \ x=-\frac{12}{2}\]


\[\large x=3\ \text{or} \ \ x=-6\]

Answer 5

Thanks a lot but...where in hell did you got the 9/4 from?

Answer 6

I took half of the x coefficient 3 to get \[\frac32\] then I squared that fraction to get \[\frac94\]. I then added this to both sides

Answer 7

This is part of what it takes to complete the square

Answer 8

thanks

Answer 9

np