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

**PythagorasNeophyte** · February 9th 2011, 10:27 PM

Hello everyone here on Math Help Forum.

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

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

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

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

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

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!

**pickslides** · February 9th 2011, 10:34 PM

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

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

Now solve for the discriminant >0

**earboth** · February 9th 2011, 10:35 PM

Originally Posted by PythagorasNeophyte

Hello everyone here on Math Help Forum.

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

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

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

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

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

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. $k^2-8k+24\ne (x-4)^2$

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

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

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

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

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.

**PythagorasNeophyte** · February 9th 2011, 10:48 PM

Thank you so much for your replies! I am very sorry for the mistake when I should have written $k$ instead of $x$ when I factorise.

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

The most important thing here is to prove that $x^2+kx = 4 -2k$ has real roots for all values of k.

**earboth** · February 9th 2011, 11:09 PM

Originally Posted by PythagorasNeophyte

Thank you so much for your replies! I am very sorry for the mistake when I should have written $k$ instead of $x$ when I factorise.

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

Never!
$(k-4)(k-4) = k^2-4k-4k+(-4)(-4)$

And I assure you $4 \cdot 4 \ne 24$

The most important thing here is to prove that $x^2+kx = 4 -2k$ 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.

**PythagorasNeophyte** · February 9th 2011, 11:24 PM

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.

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

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

What do I do with the 8?

**pickslides** · February 10th 2011, 01:45 AM

Originally Posted by pickslides

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

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

Now solve for the discriminant >0

Using what I have already said before hand

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

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

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

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

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

$\dots$

**CaptainBlack** · February 10th 2011, 08:34 PM

Originally Posted by pickslides

Using what I have already said before hand

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

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

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

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

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

$\dots$

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

CB

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