How do i put x^2+(k+3)x+4x=0 into the quadratic formula?

Question

lieutenantgeneral · Answer

x^2 + 4x + 3 = 0 
x = [-b ±√(b^2 - 4ac)]/2a 

a = 1 
b = 4 
c = 3 

x = [-4 ±√(16 - 12)]/2 
x = [-4 ±√4]/2 
x = [-4 ±2]/2 

x = [-4 + 2]/2 
x = [-2]/2 
x = -1 

x = [-4 - 2]/2 
x = [-6]/2 
x = -3 

∴ x = -3 , -1

anonymous · Answer

@LieutenantGeneral, but where did k go? 
the whole question is to find the value of k, and i have to use the quadratic formula to find it... In the book it says, I have to use d=b^2 -4ac but doesn't seem right

hartnn · Answer

is it 4x in the end ?? or just 4 ?

lieutenantgeneral · Answer

www.wolframalpha.com/
or type it in on google

anonymous · Answer

it's 4x, @hartnn

hartnn · Answer

hmm, ok
so first combine terms with x
$\Large (k+3)x +4x = (k+3+4)x = (k+7)x$
got this ?

anonymous · Answer

Ahh, I was confused on how to add the 4x to (k+3)x 
okay, got it!

hartnn · Answer

ok, now compare your equation with $ax^2+bx+c=0$

can you get values for a,b,c first ?

anonymous · Answer

yeah, so a=1, b=k+7 and c=0 
right?

hartnn · Answer

absolutely correct :)
can you proceed ?
ask if you still need help solving further.

anonymous · Answer

Yeah, i think I got it so far... I'll tag you, if i'm stuck... Thanks! :D

hartnn · Answer

welcome ^_^

anonymous · Answer

okay, i already need help... 
Is this right? 
d=(k+7)^2 -4(1)(0)
=(k^2+14k)-0 -> d=k^2+14k?

anonymous · Answer

I just noticed i got the whole question wrong!  It's 4k in the end, not 4x, @hartnn

hartnn · Answer

hahaha! i thought something was weird with the question :P

hartnn · Answer

find a,b,c values again :)

anonymous · Answer

I need to get glasses, lol :P 
okay so a=1, b=(k+3) and c=4k?

hartnn · Answer

correct!

anonymous · Answer

ahhh, see, that's what i have on my paper. but then i got confused because i wrote 4x instead of 4k.. lol 

so, d= (k+3)^2 -4(1)(4k)
=k^2 +6k+9 - 16k ? 
=k^2 -10k + 9 ?
do I factorise it and solve to get k? @hartnn

hartnn · Answer

you can factorise it but what do you actually want to find ?

anonymous · Answer

i need to find the values for k to be able to solve the equation

hartnn · Answer

you need values of "k" for which your quadratic equation has "EQUAL ROOTS" ?

hartnn · Answer

for equal roots,
d= 0
k^2-10k +9 =0 
and you get 2 integer values of k from here

anonymous · Answer

how do i get that? 
do i use the quadratic equation within the quadratic equation?

anonymous · Answer

i mean, like get two values for the square root to use in the quadratic equation ?

hartnn · Answer

so you just want to solve this : 
 x^2+(k+3)x+4x=0

and get the answer in terms of k ?

anonymous · Answer

In order to solve x^2+(k+3)x+4x=0, i have to find the values of k first... or that's what it says

hartnn · Answer

$b^2-4ac = k^2-10k+9$ 
$\Large x_{1,2} = \dfrac{-(k+3)\pm \sqrt{k^2-10k+9}}{2}$

to get the value of "k", you really need more information than just that quadratic equation

anonymous · Answer

when i translate the question to english, it says this 
A6. Determine the values ​​of k that makes the equation has exactly one solution:

I tried putting the equation into wolfamalpha.com and it shows the solutions for x : $$x = 1/2 (-\sqrt(k^2-10 k+9)-k-3)$$  and/or $$x = 1/2 (\sqrt(k^2-10 k+9)-k-3)$$  - but i don't know how, @hartnn

hartnn · Answer

YES! this is exactly what i needed!!
"Determine the values ​​of k that makes the equation has exactly one solution"

hartnn · Answer

for a quadratic equation to have equal solutions, 
$\Large  b^2-4ac = 0 $

hartnn · Answer

so, 
$k^2 - 10k + 9=0$

hartnn · Answer

so, 
$k^2-9k-k+9=0$

can you solve further by factoring method ?

anonymous · Answer

Oh, so i do need to factorise it? 
i wrote that on the paper, hold on

hartnn · Answer

yes, you get good 2 integer values of k :)

anonymous · Answer

so, 
isn't it (k-1)(k-9)=0

hartnn · Answer

yes
so what are the 2 values of k ?

anonymous · Answer

...i don't know, how do i find that? @hartnn

hartnn · Answer

$\large AB=0 \implies A=0 ~or ~ B=0 \ \large (k-1)(k-9)=0 \implies k-1=0 ~or~ k-9=0$

anonymous · Answer

Okay, so is that k=1 or k=9 ?

hartnn · Answer

yes!
thats it!

anonymous · Answer

Okay, so what do i do now? 
do i use the quadratic equation to find two solutions because the two different k values? @hartnn

hartnn · Answer

you have already completely solved your question,
"Determine the values ​​of k that makes the equation has exactly one solution"

they need values of k !
and you got them as 
k = 1 or 9

anonymous · Answer

ohhhhhhhhh, that was all? Well that easy once you knew the real question xP 
Sometimes it helps understanding the question first, haha... 
THANKS! :D Now I can hand in my assignment @hartnn

hartnn · Answer

you're welcome ^_^
good luck!

anonymous · Answer

:D Thanks, again!