SOLUTION: c)(i) Write the quadratic expression x^2-34x+9 in completed square form (ii) use the completed square form from part (c) (i) to solve the equation x^2-34x+9=0, leaving your answ

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: c)(i) Write the quadratic expression x^2-34x+9 in completed square form (ii) use the completed square form from part (c) (i) to solve the equation x^2-34x+9=0, leaving your answ      Log On


   

Question 766811: c)(i) Write the quadratic expression x^2-34x+9 in completed square form
(ii) use the completed square form from part (c) (i) to solve the equation x^2-34x+9=0, leaving your answer in exact (surd) form
(iii) use the completed square form from part (c)(i)to write down the vertex of the parabola y=x^2-34x+9
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
c)(i) Write the quadratic expression x%5E2-34x%2B9 in completed square form
Solved by pluggable solver: Completing the Square to Get a Quadratic into Vertex Form


y=1+x%5E2-34+x%2B9 Start with the given equation



y-9=1+x%5E2-34+x Subtract 9 from both sides



y-9=1%28x%5E2-34x%29 Factor out the leading coefficient 1



Take half of the x coefficient -34 to get -17 (ie %281%2F2%29%28-34%29=-17).


Now square -17 to get 289 (ie %28-17%29%5E2=%28-17%29%28-17%29=289)





y-9=1%28x%5E2-34x%2B289-289%29 Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of 289 does not change the equation




y-9=1%28%28x-17%29%5E2-289%29 Now factor x%5E2-34x%2B289 to get %28x-17%29%5E2



y-9=1%28x-17%29%5E2-1%28289%29 Distribute



y-9=1%28x-17%29%5E2-289 Multiply



y=1%28x-17%29%5E2-289%2B9 Now add 9 to both sides to isolate y



y=1%28x-17%29%5E2-280 Combine like terms




Now the quadratic is in vertex form y=a%28x-h%29%5E2%2Bk where a=1, h=17, and k=-280. Remember (h,k) is the vertex and "a" is the stretch/compression factor.




Check:


Notice if we graph the original equation y=1x%5E2-34x%2B9 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C1x%5E2-34x%2B9%29 Graph of y=1x%5E2-34x%2B9. Notice how the vertex is (17,-280).



Notice if we graph the final equation y=1%28x-17%29%5E2-280 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C1%28x-17%29%5E2-280%29 Graph of y=1%28x-17%29%5E2-280. Notice how the vertex is also (17,-280).



So if these two equations were graphed on the same coordinate plane, one would overlap another perfectly. So this visually verifies our answer.





(ii) use the completed square form from part (c) (i) to solve the equation x%5E2-34x%2B9=0, leaving your answer in exact (surd) form
%28x-17%29%5E2-280+=+0
(iii) use the completed square form from part (c)(i)to write down the vertex of the parabola y=x%5E2-34x%2B9
y=%28x-17%29%5E2-280+....compare to y=%28x-h%29%5E2%2Bk and you will see that h=17 and k=-280 which means
the vertex of the parabola is at (17,-280

since the graph above is not good, I will make another one for you to see it is a parabola:

+graph%28+600%2C+600%2C+-50%2C+50%2C+-330%2C+50%2C%28x-17%29%5E2-280+%29+