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Math Help - Proving there are no roots to a quadratic equation

  1. #1
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    Proving there are no roots to a quadratic equation

    Hello everyone,

    I understand perfectly how to solve quadratic equations, but have come across a question which somewhat confuses me. I hope someone can help:

    "Show that any root of the equation 5 + x - sqrt(3 + 4x) = 0 is also a root of the equation x^2 + 6x + 22 = 0. Hence show that the equation 5 + x - sqrt(3 + 4x) = 0 has no solutions."

    What I don't understand is that it is obvious eq1 and eq2 have no solutions, so how can I possibly show that any roots of eq1 are also roots of eq2?

    Thank you.
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  2. #2
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    Re: Proving there are no roots to a quadratic equation

    Quote Originally Posted by mabentley View Post
    Hello everyone,

    I understand perfectly how to solve quadratic equations, but have come across a question which somewhat confuses me. I hope someone can help:

    "Show that any root of the equation 5 + x - sqrt(3 + 4x) = 0 is also a root of the equation x^2 + 6x + 22 = 0. Hence show that the equation 5 + x - sqrt(3 + 4x) = 0 has no solutions."

    What I don't understand is that it is obvious eq1 and eq2 have no solutions, so how can I possibly show that any roots of eq1 are also roots of eq2?

    Thank you.
    5+x-\sqrt{3+4x} \Rightarrow 5+x=\sqrt{3+4x}. Square both the sides. What do you get?
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  3. #3
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    Re: Proving there are no roots to a quadratic equation

    Thanks for the reply.

    I can see that by squaring the original formula I get a quadratic:

    x^2 - 4x + 22 = 0

    But the question asks, "Show that any root of this equation...". I can't see that this equation has any roots...
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  4. #4
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    Re: Proving there are no roots to a quadratic equation

    Quote Originally Posted by mabentley View Post
    I can see that by squaring the original formula I get a quadratic:
    x^2 - 4x + 22 = 0
    But the question asks, "Show that any root of this equation...". I can't see that this equation has any roots...
    Frankly, I find this to be a very odd question.
    I think whoever set the question has a very specific answer in mind.
    Therefore, you are best served by asking your instructor about it.
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  5. #5
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    Re: Proving there are no roots to a quadratic equation

    Thanks, I'm glad someone agrees. I'm looking back over some A-Level maths (not done any in 7 years) and found this question in the 'past papers' part of an old book. I'm generally finding the questions quite easy, but I'm struggling to see what is wanted from me for this question. I'll give it a miss unless anyone knows better...
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    Senior Member abhishekkgp's Avatar
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    Re: Proving there are no roots to a quadratic equation

    Quote Originally Posted by mabentley View Post
    Thanks for the reply.

    I can see that by squaring the original formula I get a quadratic:

    x^2 - 4x + 22 = 0

    But the question asks, "Show that any root of this equation...". I can't see that this equation has any roots...
    I think you committed a mistake there.
    what i did was
    5+x=\sqrt{3+4x} \Rightarrow (x+5)^2=3+4x \Rightarrow x^2+25+10x=3+4x \Rightarrow x^2+(10-4)x + 25-3=0 \Rightarrow x^2+6x+22=0.
    Now do you get it?
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  7. #7
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    Re: Proving there are no roots to a quadratic equation

    Quote Originally Posted by mabentley View Post
    Thanks for the reply.

    I can see that by squaring the original formula I get a quadratic:

    x^2 - 4x + 22 = 0

    But the question asks, "Show that any root of this equation...". I can't see that this equation has any roots...
    I don't get that...

    \displaystyle \begin{align*} 5 + x &= \sqrt{3 + 4x} \\ (5 + x)^2 &= \left(\sqrt{3 + 4x}\right)^2 \\ 25 + 10x + x^2 &= 3 + 4x \\ x^2 + 6x + 22 &= 0  \end{align*}

    which is the quadratic you were given...
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  8. #8
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    Re: Proving there are no roots to a quadratic equation

    Quote Originally Posted by abhishekkgp View Post
    I think you committed a mistake there.
    what i did was
    5+x=\sqrt{3+4x} \Rightarrow (x+5)^2=3+4x \Rightarrow x^2+25+10x=3+4x \Rightarrow x^2+(10-4)x + 25-3=0 \Rightarrow x^2+6x+22=0.
    Now do you get it?
    Oops, you are right. That was very sloppy of me - I squared each individual term rather than squaring both sides. It makes complete sense now.
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