# 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
The content structure of the passage involves a student asking for help with a math problem and receiving guidance from other forum members. The student is trying to prove that a given quadratic equation has real roots for all real values of the parameter \( k \).
single_question_with_answer

the student's question is about showing that the equation \( x^2 + kx = 4 - 2k \) has real roots for all real values of \( k \). the answer is provided through the discussion, where the student and forum members work through the problem, ultimately showing that the discriminant \( k^2 - 8k + 24 \) is always non-negative, which proves the equation has real roots for all real values of \( k \).
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# 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
user
The content structure of the passage involves a student asking for help with a math problem and receiving guidance from other forum members. The student is trying to prove that a given quadratic equation has real roots for all real values of the parameter \( k \).

TYPE: SINGLE_QUESTION_WITH_ANSWER

The student's question is about showing that the equation \( x^2 + kx = 4 - 2k \) has real roots for all real values of \( k \). The answer is provided through the discussion, where the student and forum members work through the problem, ultimately showing that the discriminant \( k^2 - 8k + 24 \) is always non-negative, which proves the equation has real roots for all real values of \( k \).