Unit 9 - HW #74 Solving Quadratics; Review for Test
12 comments:
Anonymous
said...
um, this question is actually about #27 on page 544, which was the hw from june 1st. I used the quadratic formula to gt the "roots"and i got x= , 4, 6 , after i put it into the parenthesis and solved for x to make them ture(come out to zero.)Then i used the formula for vertexes and I got 3 1/2,O. Because my computer wouldn't let me plug that into the graphing calculator, i checked my answer in the back of the book and it just said "3,4." What did I do wrong and how do I go about getting that answer?
This problem might be a little confusing becuz it uses "y" for the independent variable (which is normally "x"). That's an OK thing for an algebra-tician to do, but for beginners it can be a little confusing. The authors of the book should be tarred and feathered! Yet, we push on...
First thing, whether you try factoring or the quadratic formula, you must put the equation in standard form, which means you need to set it equal to zero. A LOT OF STUDENTS WILL MISS THIS STEP. Before you start, you must add 6 to both sides, so:
y^2 -7y + 6 = -6 ... becomes
y^2 -7y + 12 = 0
So, a=1; b=-7 and c=12
This is a great candidate for factoring (the leading coefficient = 1 and the other coefficients are integers), and gosh (!!) we find that the two numbers that multiply to 12 with a sum of -7 are (drum roll please) -3 and -4. So, your factored equation looks like this:
(y-3)(y-4)=0
Now, which value or values of y will make this equation true? The ZERO PRODUCT RULE leads you clearly to a solution of y = 3,4.
The quadratic formula is just an alternate method of finding the solution(s). It WILL ALWAYS yield the SAME RESULT. Re-check your work. The axis of symmetry should always split the difference, so 3 1/2 is correct... why you also list a 0, I don't know. The axis of symmetry can only "reside" at a single location. Further, this problem doesn't ask you for an axis of symmetry, so there is no need to bother with it.
If you were to graph x^2 - 7x + 12 on a graphing calculator (borrow your brother's or go to the library!), you would see a parabola, opening up, with roots (x-intercepts) at 3,4. Ca-peesh??
And get a working PC that can handle Adobe PDFs, Java, and Flash. The folks at the high school will be using these tools, too!
okay...so i got really excited and thought that i was doing really well at this unit but i couldn't be more wrong.
on page 499 numbers 13-16 how do you match the equations with the graphs????
also, i looked again at the problem set 9B and had trouble assigning numbers to the variables...can sum1 tell me what the assigned numbers are and a brief description why? thanks
okay, pg. 499, 13-16. So let's go to a simpler problem. What can we tell about this equation? y = x^2 + x + 0 Well. We can tell that it opens up. Also we can tell that is is a "normal" quadratic since it doesn't go really fast or slow. Also, don't forget x=-b/2a, which you can use to find the vertex.
Finally, if using this doesn't give you enough just use the quadratic equation. The quadratic equation gives us the roots of the problem. So, you can find the roots, (where the parabola crosses the x axis) and where the vertex is.
Also, if you want to double check just use the graphing calculator.
We should know open-up vs. open-down (leading coefficient is positive vs. negative, respectively). We should also be able to locate y-intercepts... what is the value of y when x=0? These two facts alone can answer these questions quickly, no need for solving, factoring, quadratic formula-ling or anything!
with the formula x=-b/2a i know that you get what the x coordinate is for the vertex, and then you plug that in for x to find out the y coordinate for the vertex.
Well, x=-b/2a definitely works EVERY TIME. So, the question is, where did you go wrong.
My experience is that most students have trouble when b is a negative number, since -b (the opposite of b, right?), will then be a positive number.
All I can say is re-check your calculation with a fine-tooth comb. There isn't that much to it. ALSO, make sure you quadratic equation is in standard form (either "y=" or "0=").
for page 544 number 27 i didnt get the right answer either. i got something totally different though and i thought i did it right. but apparently i didnt. i got x=-1,-6 so i am really confused. HELP!
Good job on pg 544 #27, anonymous! The secret (shhh!) is that the equation needs to be in standard form, i.e. set try factoring, complete the square, or use the quadratic formula. What fun!
David, to complete the square for:
x^2 - 8x + 11 = 0
... this is what you need to do.
Re-write the equation, without a "3rd term" by subtracting 11 fbs...
x^2 - 8x + = -11
Then, what would make the left side a perfect square trinomial? The number you need is half of -8 squared, which is -4 squared, otherwise known as 16! You'll have to add 16 tbs.
x^2 - 8x + 16 = -11 + 16
so, now you have...
(x - 4)^2 = 5
take the square root of both sides...
x - 4 = +/- sqrt(5)
solve for x by adding 4 tbs...
x = 4 +/- sqrt(5) x = 4 +/- 2.24 x = 1.76, 6.24
Ca-peesh??!!
The other method would be to take the original equation (in standard form, of course) and find your a,b,c and use the quadratic formula. Try it and see, you should (you MUST) get the same answer!
12 comments:
um, this question is actually about #27 on page 544, which was the hw from june 1st. I used the quadratic formula to gt the "roots"and i got x= , 4, 6 , after i put it into the parenthesis and solved for x to make them ture(come out to zero.)Then i used the formula for vertexes and I got 3 1/2,O. Because my computer wouldn't let me plug that into the graphing calculator, i checked my answer in the back of the book and it just said "3,4." What did I do wrong and how do I go about getting that answer?
This problem might be a little confusing becuz it uses "y" for the independent variable (which is normally "x"). That's an OK thing for an algebra-tician to do, but for beginners it can be a little confusing. The authors of the book should be tarred and feathered! Yet, we push on...
First thing, whether you try factoring or the quadratic formula, you must put the equation in standard form, which means you need to set it equal to zero. A LOT OF STUDENTS WILL MISS THIS STEP. Before you start, you must add 6 to both sides, so:
y^2 -7y + 6 = -6 ... becomes
y^2 -7y + 12 = 0
So, a=1; b=-7 and c=12
This is a great candidate for factoring (the leading coefficient = 1 and the other coefficients are integers), and gosh (!!) we find that the two numbers that multiply to 12 with a sum of -7 are (drum roll please) -3 and -4. So, your factored equation looks like this:
(y-3)(y-4)=0
Now, which value or values of y will make this equation true? The ZERO PRODUCT RULE leads you clearly to a solution of y = 3,4.
The quadratic formula is just an alternate method of finding the solution(s). It WILL ALWAYS yield the SAME RESULT. Re-check your work. The axis of symmetry should always split the difference, so 3 1/2 is correct... why you also list a 0, I don't know. The axis of symmetry can only "reside" at a single location. Further, this problem doesn't ask you for an axis of symmetry, so there is no need to bother with it.
If you were to graph x^2 - 7x + 12 on a graphing calculator (borrow your brother's or go to the library!), you would see a parabola, opening up, with roots (x-intercepts) at 3,4. Ca-peesh??
And get a working PC that can handle Adobe PDFs, Java, and Flash. The folks at the high school will be using these tools, too!
okay...so i got really excited and thought that i was doing really well at this unit but i couldn't be more wrong.
on page 499 numbers 13-16 how do you match the equations with the graphs????
also, i looked again at the problem set 9B and had trouble assigning numbers to the variables...can sum1 tell me what the assigned numbers are and a brief description why? thanks
okay, pg. 499, 13-16. So let's go to a simpler problem.
What can we tell about this equation?
y = x^2 + x + 0
Well. We can tell that it opens up. Also we can tell that is is a "normal" quadratic since it doesn't go really fast or slow.
Also, don't forget x=-b/2a, which you can use to find the vertex.
Finally, if using this doesn't give you enough just use the quadratic equation. The quadratic equation gives us the roots of the problem. So, you can find the roots, (where the parabola crosses the x axis) and where the vertex is.
Also, if you want to double check just use the graphing calculator.
Sorry, it posted the same comment three times.
BUT, with the problem set. What do you mean by assigning numbers to variables?
Soval is so very correct.
We should know open-up vs. open-down (leading coefficient is positive vs. negative, respectively). We should also be able to locate y-intercepts... what is the value of y when x=0? These two facts alone can answer these questions quickly, no need for solving, factoring, quadratic formula-ling or anything!
okay...so im having more trouble =[
with the formula x=-b/2a
i know that you get what the x coordinate is for the vertex, and then you plug that in for x to find out the y coordinate for the vertex.
why dosent this work?????
Well, x=-b/2a definitely works EVERY TIME. So, the question is, where did you go wrong.
My experience is that most students have trouble when b is a negative number, since -b (the opposite of b, right?), will then be a positive number.
All I can say is re-check your calculation with a fine-tooth comb. There isn't that much to it. ALSO, make sure you quadratic equation is in standard form (either "y=" or "0=").
Mr. C
Complete the square...I have a hard time doing this. For example how would you solve..
x^2-8x+11=0 ?
for page 544 number 27 i didnt get the right answer either. i got something totally different though and i thought i did it right. but apparently i didnt. i got x=-1,-6 so i am really confused. HELP!
never mind. i realized that i had to turn the -6 into 0 and then i got the correct answer. thanks though.
Good job on pg 544 #27, anonymous! The secret (shhh!) is that the equation needs to be in standard form, i.e. set try factoring, complete the square, or use the quadratic formula. What fun!
David, to complete the square for:
x^2 - 8x + 11 = 0
... this is what you need to do.
Re-write the equation, without a "3rd term" by subtracting 11 fbs...
x^2 - 8x + = -11
Then, what would make the left side a perfect square trinomial? The number you need is half of -8 squared, which is -4 squared, otherwise known as 16!
You'll have to add 16 tbs.
x^2 - 8x + 16 = -11 + 16
so, now you have...
(x - 4)^2 = 5
take the square root of both sides...
x - 4 = +/- sqrt(5)
solve for x by adding 4 tbs...
x = 4 +/- sqrt(5)
x = 4 +/- 2.24
x = 1.76, 6.24
Ca-peesh??!!
The other method would be to take the original equation (in standard form, of course) and find your a,b,c and use the quadratic formula. Try it and see, you should (you MUST) get the same answer!
a= 1
b= - 8
c= 11
Post a Comment