What you have is (at least to me) a rather unorthodox method of a process called "completing the square." Despite the very unfamiliar process, I have proven to myself that it actually works in the general case.
Start with the general form of the quadratic equation:
Now, let's follow your steps:
"a. move the constant term to the right side of the equation"
"b. multiply each term in the equation by four times the cofficient of the
"c. square the coefficient of the original x term and add it to both sides of the equation"
"d. take the square root of both sides"
First you need to factor the expression on the LHS of the equation. Fortunately, this is a perfect square expression:
(Verification of that last intermediate step is left as an exercise for the student)
Now we can take the square root of both sides, remembering to consider both the positive and negative roots -- such consideration typically indicated in the resulting RHS.
By considering both positive and negative roots at this point, you can do steps e and f simultaneously:
And we achieve a result that is identical to the quadratic formula which is the solution to the general quadratic.
Let's do one of your problems now that we are certain that the process will work for every problem:
a:
b:
c:
d:
d1:
e&f:
Then you can simplify a little by recognizing that 56 is 4 times 14, so:
Let's see if this answer makes any sense. 3 squared is 9 and 4 squared is 16, so 3.75 is a pretty good rough guess for square root 14. 1 plus 3.75 is 4.75 and 1 minus 3.75 is -2.75. So look where the graph crosses the
Pretty close, I'd say. Let me know if you still need help.
John