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Thread: Use the Quadratic Formula to solve: x^2 10x 1 = -10

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    Use the Quadratic Formula to solve: x^2 10x 1 = -10

    Use the Quadratic Formula to solve:
    x^2 10x 1 = -10
    x^2 10x 1+10 = -10 +10
    x^2 10x + 9 = 0

    a=1 b=10 c=9

    x= [--10 ?-10^2 - 4(1)(9)] / 2(1)
    x= [10 ?100 + 36] / 2
    x= [10 ?136] / 2
    x= [10 ?8?17] / 2
    x= [10 2?17] / 2
    x= 5 2?17

    Did I simplify 136 correctly? Please let me know if anything is off here! I appreciate it!

    Flora

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    Re: Use the Quadratic Formula to solve: x^2 10x 1 = -10

    Quote Originally Posted by flora33
    x= [--10 ?-10^2 - 4(1)(9)] / 2(1)
    x= [10 ?100 + 36] / 2
    x= [10 ?136] / 2
    x= [10 ?8?17] / 2
    x= [10 2?17] / 2
    x= 5 2?17

    Did I simplify 136 correctly? Please let me know if anything is off here! I appreciate it!
    I'm afraid you have an arithmetic mistake.

    (-10)^2 - 4(1)(9) should be

    100 - 36

    So, under the radical sign you should have 64 rather than 136.

    Try that....

    And then CHECK your answer (if you'd checked the answer you got, you would already know it is incorrect).

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    Re: Use the Quadratic Formula to solve: x^2 10x 1 = -10

    Quote Originally Posted by Mrspi
    I'm afraid you have an arithmetic mistake.

    (-10)^2 - 4(1)(9) should be

    100 - 36

    So, under the radical sign you should have 64 rather than 136.

    Try that....And then CHECK your answer....
    I would have checked my answer, and this may sound ridiculous, but I don't know how to put an answer with a radical symbol back in to the original equation to check it. I tried to ask my teacher in "class" which was just a live chat we have once a week and all he said was, "plug in the numbers" and moved on. He's pretty good at ignoring questions. I tried to look at how the book did it within the examples and when I "plugged the numbers" in, I didn't get the same answers the book did... So, anyway, that is that. If you want to tell me how I can check that using my calculator that would be awesome!

    I went through the problem again and here is what I've come up with (I did check and these seem right to me):
    x= [10 ?100 - 36] / 2
    x= [10 ?64] / 2
    x= [10 ?8?8] / 2
    x= [10 8] / 2
    x= 10 + 8 / 2 = 9
    x= 10 - 8 / 2 = 1

    Thanks for your help!

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    Re: Use the Quadratic Formula to solve: x^2 10x 1 = -10

    oops!

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    Elite Member stapel's Avatar
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    Quote Originally Posted by flora33
    I don't know how to put an answer with a radical symbol back in to the original equation to check it.
    There is no separate or "different" process for plug-n-chug if the plug-in value happens to contain radicals. You still plug the value, radicals and all, in for the variable, and simplify.

    If, for instance, the quadratic equation were "x[sup:1wloantl]2[/sup:1wloantl] - 2x - 4 = 0", so the solution were x = 1 +/- sqrt[5], you'd plug 1 - sqrt[5] in for x and simplify, and then you'd plug 1 + sqrt[5] in for x and simplify. The intermediate steps of the simplification might be a bit "lumpier", but otherwise the process will be exactly the same as for any other x-value.

    If I have misunderstood your meaning, kindly reply with clarification. Thank you!

    Eliz.

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    Re: Re:

    Quote Originally Posted by stapel
    If, for instance, the quadratic equation were "x[sup:33efbcgt]2[/sup:33efbcgt] - 2x - 4 = 0", so the solution were x = 1 +/- sqrt[5], you'd plug 1 - sqrt[5] in for x and simplify, and then you'd plug 1 + sqrt[5] in for x and simplify. The intermediate steps of the simplification might be a bit "lumpier", but otherwise the process will be exactly the same as for any other x-value.
    So I would do, for your example 1 - 25 = 24 because 5 squared is 25, right? I tried this on another problem which I'm having a hard time with:
    x^2 3x = -6x 1
    x^2 + 3x + 1 = 0
    a = 1 b = 3 c = 1
    x = -3 ?[3^2 - 4(1)(1)] / 2(1)
    x = -3 ?[9 - 4] / 2
    x = -3 ?[5] / 2 So is this one answer for x, or do I further simplify by doing:
    x = -3 + ?[5] / 2 = 22 / 2 = 11
    x = -3 - ?[5] / 2 = -28 / 2 = -14
    (-3 + ?[5] / 2)^2 + 3(-3 + ?[5] / 2) + 1 = 0 (I got 94.75)
    (-3 - ?[5] / 2)^2 + 3(-3 - ?[5] / 2) + 1 = 0 (I got 219.75)
    or should I have used the 11 and -14:
    11^2 + 3(11) + 1 = 0 (I got 155)
    -14^2 + 3(-14) + 1 = 0 (I got 155)

    OK! I guess I answered my own question here... but, I still am curious about your example, x = 1 +/- sqrt[5]. Thanks!

    Flora

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    Elite Member stapel's Avatar
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    Quote Originally Posted by flora33
    So I would do, for your example 1 - 25 = 24 because 5 squared is 25, right?
    Um... what...? Where is "25" coming from? And how is 1 - 25 not negative?

    You have "x[sup:2kyx8331]2[/sup:2kyx8331] - 2x - 4 = 0", and a proposed solution value of x = 1 - sqrt[5]. To "plug in for x", you would take the "1 - sqrt[5]" and use this to replace (plug in for) every instance of "x" in the equation:

    . . . . .(1 - sqrt[5])[sup:2kyx8331]2[/sup:2kyx8331] - 2(1 - sqrt[5]) - 4 ?=? 0

    You would then square out the first term:

    . . . . .(1 - 2 sqrt[5] + 5) - 2(1 - sqrt[5]) - 4 ?=? 0

    ...and simplify:

    . . . . .1 + 5 - 2 sqrt[5] - 2 + 2 sqrt[5] - 4 ?=? 0

    . . . . .1 + 5 - 2 - 4 - 2 sqrt[5] + 2 sqrt[5] ?=? 0

    . . . . .6 - 6 + 2 sqrt[5] - 2 sqrt[5] ?=? 0

    . . . . .0 + 0 = 0

    See? It works just like every other plug-n-chug or check-the-solution you've done. No new rules; no new formulas or processes; just the exact same plugging in and checking as always.

    Eliz.

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    Re: Use the Quadratic Formula to solve: x^2 10x 1 = -10

    Quote Originally Posted by flora33

    I would have checked my answer, and this may sound ridiculous, but I don't know how to put an answer with a radical symbol back in to the original equation to check it. I tried to ask my teacher in "class" which was just a live chat we have once a week and all he said was, "plug in the numbers" and moved on. He's pretty good at ignoring questions. I tried to look at how the book did it within the examples and when I "plugged the numbers" in, I didn't get the same answers the book did... So, anyway, that is that. If you want to tell me how I can check that using my calculator that would be awesome!

    !

    Yet ANOTHER reason why I do not believe that an "online class" is of any benefit.

    If you are PAYING for this class, you are being cheated of real instruction.

    PLEASE check to see if there is a community or junior college in your area which might offer you a real class, with a real teacher, and the opportunity to interact with your classmates and schedule actual face-to-face office time with your instructor. As you are no doubt finding out (the hard way, it seems) online instruction is surely not for everyone.

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    Re: Use the Quadratic Formula to solve: x^2 10x 1 = -10

    Quote Originally Posted by Mrspi
    Quote Originally Posted by flora33

    I would have checked my answer, and this may sound ridiculous, but I don't know how to put an answer with a radical symbol back in to the original equation to check it. I tried to ask my teacher in "class" which was just a live chat we have once a week and all he said was, "plug in the numbers" and moved on. He's pretty good at ignoring questions. I tried to look at how the book did it within the examples and when I "plugged the numbers" in, I didn't get the same answers the book did... So, anyway, that is that. If you want to tell me how I can check that using my calculator that would be awesome!

    !

    Yet ANOTHER reason why I do not believe that an "online class" is of any benefit.

    If you are PAYING for this class, you are being cheated of real instruction.

    PLEASE check to see if there is a community or junior college in your area which might offer you a real class, with a real teacher, and the opportunity to interact with your classmates and schedule actual face-to-face office time with your instructor. As you are no doubt finding out (the hard way, it seems) online instruction is surely not for everyone.
    Well, thank you for your advice but I have 2 weeks left in this class so obviously I'm not quitting now. My family moves every year, so it wouldn't make sense to take classes locally because then, I am probably really going to cheat myself... out of credits that don't transfer to the next school I go to. I really don't agree that "online classes" are of no benefit. I would say I've learned a lot more from this class then I might have in real time because I've been FORCED to spend a lot of time at it! I am also a stay at home mother to a small baby, so I enjoy online classes, and they ARE for me... What is not for me, is algebra. Yeah, I've had a hard time but I'm not unintelligent. It may seem that way since I'm posting here constantly, but I'm just making sure I understand everything as best as possible before completing my work for a grade. I appreciate all the help and insight I receive here though I'll sure be glad when I don't have to think about algebra anymore!

    Flora

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    Re:

    Quote Originally Posted by stapel
    Quote Originally Posted by flora33
    So I would do, for your example 1 - 25 = 24 because 5 squared is 25, right?
    Um... what...? Where is "25" coming from? And how is 1 - 25 not negative?

    You have "x[sup:28epy69h]2[/sup:28epy69h] - 2x - 4 = 0", and a proposed solution value of x = 1 - sqrt[5]. To "plug in for x", you would take the "1 - sqrt[5]" and use this to replace (plug in for) every instance of "x" in the equation:

    . . . . .(1 - sqrt[5])[sup:28epy69h]2[/sup:28epy69h] - 2(1 - sqrt[5]) - 4 ?=? 0

    You would then square out the first term:

    . . . . .(1 - 2 sqrt[5] + 5) - 2(1 - sqrt[5]) - 4 ?=? 0

    ...and simplify:

    . . . . .1 + 5 - 2 sqrt[5] - 2 + 2 sqrt[5] - 4 ?=? 0

    . . . . .1 + 5 - 2 - 4 - 2 sqrt[5] + 2 sqrt[5] ?=? 0

    . . . . .6 - 6 + 2 sqrt[5] - 2 sqrt[5] ?=? 0

    . . . . .0 + 0 = 0

    See? It works just like every other plug-n-chug or check-the-solution you've done. No new rules; no new formulas or processes; just the exact same plugging in and checking as always.

    Eliz.
    All right, I see what you mean. Thanks for your help.

    Flora

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