text: # Showing that the equation has real roots for all real values of 'k'
  
  • Feb 9th 2011, 10:27 PM
  PythagorasNeophyte
  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!
  • Feb 9th 2011, 10:34 PM
  pickslides
  $\displaystyle x^2 + kx - 4 + 2k =0$
  
  $\displaystyle a=1, b=k, c= - 4 + 2k$
  
  Now solve for the discriminant >0
  • Feb 9th 2011, 10:35 PM
  earboth
  Quote:
  
  Originally Posted by PythagorasNeophyte
  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.
  • Feb 9th 2011, 10:48 PM
  PythagorasNeophyte
  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.
  • Feb 9th 2011, 11:09 PM
  earboth
  Quote:
  
  Originally Posted by PythagorasNeophyte
  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 $\displaystyle 4 \cdot 4 \ne 24$
  Quote:
  
  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.
  • Feb 9th 2011, 11:24 PM
  PythagorasNeophyte
  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?
  • Feb 10th 2011, 01:45 AM
  pickslides
  Quote:
  
  Originally Posted by pickslides
  $\displaystyle x^2 + kx - 4 + 2k =0$
  
  $\displaystyle a=1, b=k, c= - 4 + 2k$
  
  Now solve for the discriminant >0
  
  Using what I have already said before hand
  
  $\displaystyle \displaystyle b^2-4ac > 0 \implies k^2 -4(1)(-4+2k)>0$
  
  $\displaystyle \displaystyle k^2 -4(1)(-4+2k)>0$
  
  $\displaystyle \displaystyle k^2 -4(-4+2k)>0$
  
  $\displaystyle \displaystyle k^2 +16-8k>0$
  
  $\displaystyle \displaystyle (k-4)^2>0$
  
  $\displaystyle \dots$
  • Feb 10th 2011, 08:34 PM
  CaptainBlack
  Quote:
  
  Originally Posted by pickslides
  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
explanation: The text extract is a math forum post where a user is seeking help to prove that a given quadratic equation has real roots for all real values of \( k \). The user provides their working and asks for corrections and further guidance. Another user, pickslides, provides a step-by-step solution to the problem, and CaptainBlack corrects a minor detail regarding the inequality.
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distilabel_metadata.role: system
distilabel_metadata.content: # Showing that the equation has real roots for all real values of 'k'
  
  • Feb 9th 2011, 10:27 PM
  PythagorasNeophyte
  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!
  • Feb 9th 2011, 10:34 PM
  pickslides
  $\displaystyle x^2 + kx - 4 + 2k =0$
  
  $\displaystyle a=1, b=k, c= - 4 + 2k$
  
  Now solve for the discriminant >0
  • Feb 9th 2011, 10:35 PM
  earboth
  Quote:
  
  Originally Posted by PythagorasNeophyte
  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.
  • Feb 9th 2011, 10:48 PM
  PythagorasNeophyte
  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.
  • Feb 9th 2011, 11:09 PM
  earboth
  Quote:
  
  Originally Posted by PythagorasNeophyte
  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 $\displaystyle 4 \cdot 4 \ne 24$
  Quote:
  
  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.
  • Feb 9th 2011, 11:24 PM
  PythagorasNeophyte
  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?
  • Feb 10th 2011, 01:45 AM
  pickslides
  Quote:
  
  Originally Posted by pickslides
  $\displaystyle x^2 + kx - 4 + 2k =0$
  
  $\displaystyle a=1, b=k, c= - 4 + 2k$
  
  Now solve for the discriminant >0
  
  Using what I have already said before hand
  
  $\displaystyle \displaystyle b^2-4ac > 0 \implies k^2 -4(1)(-4+2k)>0$
  
  $\displaystyle \displaystyle k^2 -4(1)(-4+2k)>0$
  
  $\displaystyle \displaystyle k^2 -4(-4+2k)>0$
  
  $\displaystyle \displaystyle k^2 +16-8k>0$
  
  $\displaystyle \displaystyle (k-4)^2>0$
  
  $\displaystyle \dots$
  • Feb 10th 2011, 08:34 PM
  CaptainBlack
  Quote:
  
  Originally Posted by pickslides
  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
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The text extract is a math forum post where a user is seeking help to prove that a given quadratic equation has real roots for all real values of \( k \). The user provides their working and asks for corrections and further guidance. Another user, pickslides, provides a step-by-step solution to the problem, and CaptainBlack corrects a minor detail regarding the inequality.
  
  Label: question-and-answer