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Wynwood SquareFull description
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Wynwood Square
Completing the square If we try to solve this quadratic equation by facto ring, 2
+ 6 x 6 x + 2
=0
we cannot. Therefore, we will will complete complete the square. We will will make the quadratic into the form 2
2
2
a + 2ab 2ab + + b
= (a + b) . 2
This technique is valid only when t he coefficient of x of x is 1. 1) Transpose the constant term to the right 2
x + 6 x 6 x = = −2. 2) Add a square number to both sides -- add the square of half the coefficient of x of x.. In this case, add the square of half of 6; that is, add the square of 3, which is 9: 2
x + 6 x 6 x + + 9 = −2 + 9. The left-hand left-hand side is now the perfect square of ( x + x + 3). 2
( x + x + 3) = 7. 3 is half of the coefficient 6. That equation has the form a2
= b
which implies a = ±
.
Therefore, x + x + 3 = ±
x = −3 ± That is, the solutions to 2
x + 6 x 6 x + 2 = 0 are the conjugate pair ,
.
−3 +
, −3 −
.
For a method of checking these roots, see the theorem of the sum and product of the roots: Lesson 10 of Topics in Precalculus, In Lesson 18 there are examples and problems in which the coefficient of x is odd. Also, some of the quadratics below have complex roots, and some involve simplifying radicals. Problem 6. Solve each quadratic equation by completing the square. To see the answer, pass your cursor from left to right over the colored area. To cover the answer again, click "Refresh" ("Reload"). Do the problem yourself first!
a)
2
− 2 x − 2 = 0
b)
2
− 10 x + 20 = 0
2
2
x − 2 x = 2
2
x − 10 x = −20
2
− 2 x + 1 = 2 + 1
− 10 x + 25 = −20 + 25
( x − 1)2 = 3
( x − 5)2 = 5
x − 1 = ±
x − 5 = ±
x = 1 ± c)
2
− 4 x + 13 = 0
2
x − 4 x = −13
x = 5 ± d)
2
+ 6 x + 29 = 0
2
x + 6 x = −29
2
2
x − 4 x + 4 = −13 + 4
x + 6 x + 9 = −29+ 9
2
( x + 3) = −20
2
( x − 2) = −9
x − 2 = ±3i
x + 3 = ±
x = 2 ± 3i e)
2
− 5 x − 5
= 0
x = −3 ± 2i f)
2
x2 − 5 x = 5
2
− 5 x + 25/4 = 5 + 25/4
2
( x − 5/2) = 5 + 25/4
x − 5/2 = ±
x =
+ 3 x + 1
= 0
x2 + 3 x = −1
2
+ 3 x + 9/4 = −1 + 9/4
2
( x + 3/2) = − 1 + 9/4
x + 3/2 = ±
x =
Proof of the quadratic formula 2
To prove the quadratic formula, we complete the square. But to do that, the coefficient of x must be 1. Therefore, we will divide both sides of the original equation by a:
on multiplying both c and a by 4a, thus making the denominators the same