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Thread: Showing that the equation has real roots for all real values of 'k'

  1. #1
    Junior Member PythagorasNeophyte's Avatar
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    Question Showing that the equation has real roots for all real values of 'k'

    Hello everyone here on Math Help Forum.

    I have an urgent question here: show that the equation $\displaystyle x^2 + kx = 4 - 2k $ has real roots for all real values of $\displaystyle k$.

    What I know is that the term ''real roots'' imply that $\displaystyle b^2 - 4ac$ is equal or more than $\displaystyle 0$. But my primal problem is that I have difficulties in showing and proving. So here are my workings:

    $\displaystyle x^2 + kx = 4 - 2k $
    $\displaystyle x^2 + kx -4 +2k = 0$

    Here $\displaystyle a = 1, b = k, c = -4 + 2k$

    $\displaystyle b^2 -4ac = k^2 - 4(1)(-4+2k)$
    $\displaystyle k^2 -8k +24$
    $\displaystyle (x-4)(x-4)$
    $\displaystyle (x-4)^2$

    I have tried my best. Please pinpoint my mistakes and kindly continue the rest of the question. Thank you so much for your help and have a nice day!
    Last edited by PythagorasNeophyte; Feb 9th 2011 at 10:49 PM.
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  2. #2
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    $\displaystyle x^2 + kx - 4 + 2k =0$

    $\displaystyle a=1, b=k, c= - 4 + 2k $

    Now solve for the discriminant >0
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  3. #3
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    Quote Originally Posted by PythagorasNeophyte View Post
    Hello everyone here on Math Help Forum.

    I have an urgent question here: show that the equation $\displaystyle x^2 + kx = 4 - 2k $has real roots for all real values of $\displaystyle k$.

    What I know is that the term ''real roots'' imply that $\displaystyle b^2 - 4ac$ is equal or more than $\displaystyle 0$. But my primal problem is that I have difficulties in showing and proving. So here are my workings:

    $\displaystyle x^2 + kx = 4 - 2k $
    $\displaystyle x^2 + kx -4 +2k = 0$

    Here $\displaystyle a = 1, b = k, c = -4 + 2k$

    $\displaystyle b^2 -4ac = k^2 - 4(1)(-4+2k)$
    $\displaystyle k^2 -8k +24$
    $\displaystyle (x-4)(x-4)$
    $\displaystyle (x-4)^2$

    I have tried my best. Please pinpoint my mistakes and continue the rest of the question. Thank you so much for your help and have a nice day!
    1. $\displaystyle k^2-8k+24\ne (x-4)^2$

    2. According to the question you are asked to show that

    $\displaystyle k^2-8k+24\geq 0 ~for\ all ~ k \in \mathbb{R}$

    $\displaystyle k^2-8k+16+8\geq 0 ~for\ all ~ k \in \mathbb{R}$

    $\displaystyle (k-4)^2+8\geq 0 ~for\ all ~ k \in \mathbb{R}$

    3. A square is allways greater or equal zero. If you add a positive number to a square the sum must be greater than zero.
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  4. #4
    Junior Member PythagorasNeophyte's Avatar
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    Thank you so much for your replies! I am very sorry for the mistake when I should have written $\displaystyle k$ instead of $\displaystyle x$ when I factorise.

    But why $\displaystyle k^2-8k+24$, when factorised, gives $\displaystyle (k-4)^2 + 8$? If I were to factorise, it will be $\displaystyle (k-4)(k-4)$; expanding $\displaystyle (k-4)(k-4) $ gives $\displaystyle k^2-8k+24$.

    The most important thing here is to prove that $\displaystyle x^2+kx = 4 -2k$ has real roots for all values of k.
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  5. #5
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    Quote Originally Posted by PythagorasNeophyte View Post
    Thank you so much for your replies! I am very sorry for the mistake when I should have written $\displaystyle k$ instead of $\displaystyle x$ when I factorise.

    But why $\displaystyle k^2-8k+24$, when factorised, gives $\displaystyle (k-4)^2 + 8$? If I were to factorise, it will be $\displaystyle (k-4)(k-4)$; expanding $\displaystyle (k-4)(k-4) $ gives $\displaystyle k^2-8k+24$.
    Never!
    $\displaystyle (k-4)(k-4) = k^2-4k-4k+(-4)(-4)$

    And I assure you $\displaystyle 4 \cdot 4 \ne 24$
    The most important thing here is to prove that $\displaystyle x^2+kx = 4 -2k$ has real roots for all values of k.
    If the discriminant - that's what you have calculated - is greater or equal to zero then there exist real roots. Compare your own text at the beginning.
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  6. #6
    Junior Member PythagorasNeophyte's Avatar
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    Oh yes, I have forgotten that I am solving an inequality. I can only factorise in that way when I am solving an equation. I am very sorry for that.

    $\displaystyle k^2 -8k +24 \geq 0$

    $\displaystyle \Rightarrow (k-4)^2 + 8 \geq 0$

    What do I do with the 8?
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    Quote Originally Posted by pickslides View Post
    $\displaystyle x^2 + kx - 4 + 2k =0$

    $\displaystyle a=1, b=k, c= - 4 + 2k $

    Now solve for the discriminant >0
    Using what I have already said before hand

    $\displaystyle \displaystyle b^2-4ac > 0 \implies k^2 -4(1)(-4+2k)>0$

    $\displaystyle \displaystyle k^2 -4(1)(-4+2k)>0$

    $\displaystyle \displaystyle k^2 -4(-4+2k)>0$

    $\displaystyle \displaystyle k^2 +16-8k>0$

    $\displaystyle \displaystyle (k-4)^2>0$

    $\displaystyle \dots $
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  8. #8
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    Quote Originally Posted by pickslides View Post
    Using what I have already said before hand

    $\displaystyle \displaystyle b^2-4ac > 0 \implies k^2 -4(1)(-4+2k)>0$

    $\displaystyle \displaystyle k^2 -4(1)(-4+2k)>0$

    $\displaystyle \displaystyle k^2 -4(-4+2k)>0$

    $\displaystyle \displaystyle k^2 +16-8k>0$

    $\displaystyle \displaystyle (k-4)^2>0$

    $\displaystyle \dots $
    You are using strict inequalities when you should be using $\displaystyle $$\ge$ (when the discriminant is zero the original polynomial has a real double root). The discriminant has a double root at $\displaystyle $$k=4$ and since the leading coefficient is positive the discriminant is always non-negative.

    CB
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