Which of the following equations has 1 + √2 as one of its roots?

Hello, everyone. I solved this one is a slightly different manner by completing the square. Remember, to complete the square, manipulate the standard form of a quadratic equation (when it is set to 0) to isolate the "C" term.

\(Ax^2+Bx+C=0\)

\(Ax^2+Bx=-C\)

To complete the square, halve the "B" coefficient and square the result, then add this value to both sides. Take (A), for example. Just remember to first push the "C" term to the right-hand side. Also, the point of completing the square is to be able to put the derived quadratic back into binomial form and then take the square root. (See step #4 below.)

\(x^2+2x=1\)

\(x^2+2x+(\frac{2}{2})^2=1+(\frac{2}{2})^2\)

\(x^2+2x+1=2\)

\((x+1)^2=2\)

\(\sqrt{(x+1)^2}=+/-\sqrt{2}\)

\(x+1=+/-\sqrt{2}\)

\(x=-1+/-\sqrt{2}\)

Of course, this does NOT match our answer, but we are on the right track.

There is no need to complete the square with (B) or (C) (although you could), since those are both perfect squares and will produce roots of 1 and -1, respectively.

\(x^2-2x+1=(x-1)^2\) (root will be 1)

\(x^2+2x+1=(x+1)^2\) (root will be -1)

Picking up with (D), we can complete the square again.

\(x^2-2x=1\)

\(x^2-2x+(\frac{-2}{2})^2=1+(\frac{-2}{2})^2\)

\(x^2-2x+1=2\)

\((x-1)^2=2\)

\(\sqrt{(x-1)^2}=+/-\sqrt{2}\)

\(x-1=+/-\sqrt{2}\)

\(x=1+/-\sqrt{2}\)

This matches our root, 1 + √2, so the answer must be (D). Sometimes completing the square can be faster than dragging out the quadratic formula, particularly when the "B" coefficient is even. (Note that if you were to complete the square with (E), you would derive a fraction, and who wants to work with those when it is not necessary?)

I hope the above method may prove useful to others. Good luck with your studies.

- Andrew