Question

i need your help @whpalmer4
solve 3x2-12x+27=0
completing the squares

Answer 1

\[3x ^{2}-12x+27=0\]
\[x ^{2}-4x+9=0,x ^{2}-4x=-9\]
\[adding both sides \left( \frac{ -4 }{2 } \right)^{2} i.e.,4\]
\[x ^{2}-4x+4=-9+4=-5\]
\[\left( x-2 \right)^{2}=5\iota ^{2}\]
\[x-2=\pm \sqrt{5}\iota \]
\[x=2\pm \sqrt{5}\iota \]

Answer 2

Okay, first thing to do when completing the square is to divide through by any coefficient of the leading term. This makes the arithmetic simpler.

\[3x^2-12x+27=0\]Divide by 3
\[x^2-4x+9=0\]Now, move the constant term to the opposite side of the equals sign
\[x^2-4x=-9\]Take half of the value of the coefficient of the non-squared term, square it, and add to both sides:
\[x^2-4x+(-4/2)^2 = -9 + (-4/2)^2\]Doing this allows us to rewrite the left side as a perfect square, \((x-2)^2\) giving us\[(x-2)^2 = -9+(-2)^2 = -5\]Take the square root of both sides and we get \[(x-2) = \pm i\sqrt{5}\]We need both the positive and negative square roots! Solve the equation for the two values of \(x\) and you are done.

Answer 3

The reason that this technique works is that a perfect square has the form\[(x+a)(x+a) = x^2+2ax+a^2\]If we manipulate our equation to just have \[x^2+2ax\] on the left, we can find the value of \(a\) by dividing the coefficient of the \(x\) term by 2, square it to create \(a^2\), and add to both sides (preserving the equality). The we take the square root to find the value of \((x+a)\) and solve for \(x\).

Answer 4

Find the positive and negative values of \((x+a)\) and solve for both values of \(x\)...