Big Ideas Math Algebra 2, 2014
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Big Ideas Math Algebra 2, 2014 View details
4. Using the Quadratic Formula
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Exercise 3 Page 121

We can always complete the square to solve a quadratic equation.

By completing the square. See solution.

Practice makes perfect
Recall that not matter what kind of solutions the quadratic equation may have, we can always complete the square. We can use this to derive a general formula for solving a quadratic equation. Let's consider the standard form of a general quadratic equation. ax^2+bx+c = 0 The first thing we need to do is to isolate the terms containing x. Then we can divide by a so that the leading coefficient becomes 1.
ax^2+bx+c = 0
ax^2+bx = - c
x^2 +b/ax = - c/a
To complete the square we should add the square of half the value of the linear term to obtain a perfect square trinomial. After doing this we can factor the perfect square trinomial as a squared binomial and proceed by calculating the square root of both sides.
x^2 +b/ax = - c/a
ax^2+bx + (b/2a)^2 = - c/a + (b/2a)^2
Rewrite LHS as a squared binomial and calculate the square root at both sides
(x + b/2a)^2 = - c/a + (b/2a)^2
(x + b/2a)^2 = - c/a + b^2/4a^2
(x + b/2a)^2 = -4ac/4a^2 + b^2/4a^2
(x + b/2a)^2 = b^()2-4ac/4a^2
x + b/2a = ± sqrt(b^()2-4ac/4a^2)
x + b/2a = ± sqrt(b^()2-4ac)/sqrt(4a^2)
x + b/2a = ± sqrt(b^()2-4ac)/sqrt(4)sqrt(a^2)
x + b/2a = ± sqrt(b^()2-4ac)/2sqrt(a^()2)
x + b/2a = ± sqrt(b^()2-4ac)/2|a|
Let's recall the definition of absolute value of a quantity.

|a| = a if a > 0 - a if a <0

Notice that considering |a| in ± sqrt(b^()2-4ac)2|a| implies leaving the signs in front of the fraction just like they are or swapping them. However, it is the same to have ± or ∓, since this only indicates that there are two possible solutions. The order does not matter. Hence, we can replace |a| with a without problems.
x + b/2a = ± sqrt(b^2-4ac)/2a
x = -b/2a ± sqrt(b^2-4ac)/2a
x = - b/2a ± sqrt(b^2-4ac)/2a
x = - b ± sqrt(b^2-4ac)/2a
The general formula we just obtained is known as The Quadratic Formula.