Week 32: Completing the square and the quadratic formula
Field Book
117 to 120
Due
Test 29 due
Method
Paper first, every step
The last new material of the year, and it is the big one. Factoring only solves the quadratics that happen to factor neatly, which is a minority of them. The quadratic formula solves every quadratic there is, factorable or not, pretty or ugly. Completing the square is where the formula comes from, which is why you learn it first even though you will use the formula more.
What you have to be able to do
Solve a quadratic by completing the square
Use the quadratic formula on any quadratic
Read a box-and-whisker plot
Solve a variation problem where the relationship involves a square
The procedures
Variation with a square
Write the relationship with k in it, putting the square where the Field Book puts it, for example y equals k times x squared.
Substitute the pair of values you were given and solve for k.
Rewrite the relationship with the number k in place, then answer the question that was actually asked.
Completing the square
Move the constant to the right side, leaving the x terms alone on the left.
Take half of the x coefficient, square it, and add that to both sides.
The left side is now a perfect square. Factor it, take the square root of both sides, and remember the plus or minus.
The quadratic formula
Write the equation as ax squared plus bx plus c equals zero, and write down a, b, and c before you substitute anything. Every time.
x equals negative b, plus or minus the square root of b squared minus 4ac, all over 2a.
Compute the part under the radical first. If it is negative there is no real solution, and that is a real answer too.
Box-and-whisker plots
Five numbers make the plot: minimum, first quartile, median, third quartile, maximum.
The box spans the first to the third quartile, which is the middle half of the data.
The whiskers reach out to the smallest and largest values.
Worked
WORKED
Solve x2 + 6x = 7 by completing the square. Half of 6 is 3, and 3 squared is 9. Add 9 to both sides: x2 + 6x + 9 = 16. The left side factors: (x + 3)2 = 16, so x + 3 = 4 or x + 3 = −4. ANSWER: x = 1 or x = −7
WORKED
Solve x2 + 5x + 6 = 0 with the quadratic formula. a = 1, b = 5, c = 6. Under the radical: 25 − 24 = 1. x = (−5 ± 1) ÷ 2. ANSWER: x = −2 or x = −3, the same answers factoring gives
Terms
Term
Meaning
completing the square
adding a term to make one side a perfect square trinomial
discriminant
the b squared minus 4ac under the radical, which tells you how many real solutions there are
median
the middle value of an ordered data set
Check yourself
1. What do you add to both sides of x2 + 8x = 5?
2. In 2x2 − 3x + 1 = 0, name a, b, and c
3. If the discriminant is negative, how many real solutions?
reveal answers
16 · a = 2, b = −3, c = 1 · none
This page is the summary, not the lesson. The full working, every step of every example, is on the board in class. Come here to remember what the week was about. Go there to learn how it is done.
-- THE UNKNOWN -- week 32, completing the square and the quadratic formulaalgebra 1 · oda