Question

solve the following equation for real values of x by completing the square. the equations are: x^2-3x+28 and 9x^2-12x+5=0

Answer 1

really? it is easiest to use formula if "middle term" is odd, but you can still complete the square if you like

Answer 2

i probably sound really clueless, but how do i complete a square?

Answer 3

like this:

Answer 4

i assume the first one is
\[x^2-3x+28=0\] yes?

Answer 5

it will have no real solutions. are you allowing complex answers?

Answer 6

yes,

any answers are allowed

Answer 7

well actually only real numbers are acceptable. if that makes any sense...

Answer 8

ok so i would use quadratic formula. but if you want to complete the square start with
\[x^2-3x=-28\]
\[(x-\frac{3}{2})^2=-28+(\frac{3}{2})^2\]
\[(x-\frac{3}{2})^2=-\frac{103}{4}\]

Answer 9

in that case stop right here because the left hand side is a perfect square and hence not negative, right hand side is negative. no solution

Answer 10

okay i see, thanks so much!

Answer 11

the next one you have to divide everything by 9 as a first step.

Answer 12

yw

Answer 13

okay i see.