How do you solve using the completing the square method #x^2-2x-10=0#?

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Answer 1 · 2016-02-29T01:57:14

See solution below.

#x^2 - 2x = 10#

#1(x^2 - 2x + m) = 10#

#m = (b/2)^2#

#m = (-2/2)^2#

#m = 1#

#1(x^2 - 2x + 1 - 1) = 10#

#1(x^2 - 2x + 1) - 1(1) = 10#

#1(x - 1)^2 = 11#

#(x - 1) = +-sqrt(11)#

#x = +-sqrt(11) + 1#

Here are a few things to remember about solving quadratic equations using the completion of square:

At step 3, you notice I give you the formula for finding the value of m, which is what will make the expression a perfect square trinomial. "b" is the middle term, with the coefficient x, not #x^2#!
To keep the equation equivalent, you must always add and subtract the value of b inside the parentheses.
When you extract the negative value of b from the parentheses, you must multiply it by the number in front of the parentheses (the number that you factored out in step 1)
Never forget that when you take the square root of a positive number you must use the #+-# sign. Failure to do so will result in only one answer, which is not enough in a quadratic equation (a quadratic equation always has two solutions)
At step 1, notice I sent the constant term (10) to the side of the equation with 0. You must always do this: the other side is just for terms a and b, which have coefficients of #x^2# and #x#, respectively.

Practice exercises:

a) #2x^2 + 8x + 15 = 0#

b) #3x^2 - 4x + 19 = -3#