1. ## Simplify and solve!!!

Solve for k in
$\displaystyle 4k-1=\sqrt{\frac{(k-2)^2+(k)^2+(k)^2+(5k-2)^2}{4}-(4k-1)^2}$

2. 1) Square both sides

2) Expand both sides fully

3) Group like terms

4) Solve.

3. $\displaystyle 4(4k-1)^2=k^2-4k+4+k^2+k^2+25k^2-10k+4-4(16k^2-8k+1)$
$\displaystyle 4(16k^2-8k+1)=k^2-4k+4+k^2+k^2+25k^2-10k+4-64k^2+32k-4$
$\displaystyle 64k^2-32k+4=k^2-4k+4+k^2+k^2+25k^2-10k+4-64k^2+32k-4$

this is my wrong attempt.

4. Originally Posted by Punch
$\displaystyle 4(4k-1)^2=k^2-4k+4+k^2+k^2+25k^2-10k+4-4(16k^2-8k+1)$
$\displaystyle 4(4k^2-8k+1)=k^2-4k+4+k^2+k^2+25k^2-10k+4-64k^2+32k-4$
$\displaystyle 16k^2-32k+4=k^2-4k+4+k^2+k^2+25k^2-10k+4-64k^2+32k-4$

this is my wrong attempt.
Wrong or just incomplete?

5. grouping them, i will get $\displaystyle -52k^2+50k=0$ which simplifies to

$\displaystyle -26k^2+25k=0$

which is wrong...

6. I get...

Originally Posted by Punch
$\displaystyle 16k^2-32k+4=k^2-4k+4+k^2+k^2+25k^2-10k+4-64k^2+32k-4$
$\displaystyle 16k^2-32k+4=-36k^2+4+18k$

$\displaystyle 20k^2-50k=0$

$\displaystyle 10k(2k-5)=0$

$\displaystyle k = 0,\frac{5}{2}$

7. from $\displaystyle 16k^2-32k+4=-36k^2+4+18k$,

$\displaystyle 16k^2+36k^2-32k-18k=0$
$\displaystyle 52k^2-50k=0$

shouldn't it be done like this?

8. I'm getting 0.2955 and -0.55919

The answer is ~0.4 (according to Microsoft Maths).

9. Originally Posted by pickslides
I get...

$\displaystyle 16k^2-32k+4=-36k^2+4+18k$

$\displaystyle 20k^2-50k=0$

$\displaystyle 10k(2k-5)=0$

$\displaystyle k = 0,\frac{5}{2}$
pickslides's answer is correct accordingly to the answer. However, I do understand his workings, let's wait for him to show us how to solve it properly.

10. ok then it's all good.

I am also wanting to know.

11. Originally Posted by Punch
$\displaystyle 4(4k-1)^2=k^2-4k+4+k^2+k^2+25k^2-10k+4-4(16k^2-8k+1)$
$\displaystyle 4(4k^2-8k+1)=k^2-4k+4+k^2+k^2+25k^2-10k+4-64k^2+32k-4$
This should be [/tex]4(16k^2- 8k+ 1)= [/tex]
$\displaystyle 16k^2-32k+4=k^2-4k+4+k^2+k^2+25k^2-10k+4-64k^2+32k-4$

this is my wrong attempt.

12. Thanks... It's getting more and more confusing... seriously need someone to show the full workings.

13. $\displaystyle 4(16k^2-8k+1)=k^2-4k+4+k^2+k^2+25k^2-10k+4-16k^2-8k+1$
$\displaystyle 64k^2-32k+4=12k^2-22k+8$
$\displaystyle 52k^2-10k-4=0$