Question

i am having problems solveing x^2-3x-1=0

Answer 1

What problem?

Answer 2

Are you using the Quadratic Formula?

Answer 3

thats what i dont understand

Answer 4

did this help?

Answer 5

meverett, it's the solution. Of course it helps.

Answer 6

ohhh ... I do not learn by someone giving me a solution ...

Answer 7

meverett, as I stated earlier, your methods are not that great.

Answer 8

yes but is that how you simplify it also how did you get it

Answer 9

You have good intentions meverett, but they are not as effective as you want them to be.

Answer 10

opps he has a question, maybe time to second think

Answer 11

Hero is using completing the square method, you can notice that L.H.S of the equation is being reduced to a square form.

Answer 12

Hero, I think you made a typo in there :)
should be x^2 - 3x + 9/4

Answer 13

and then applying square root gives the solution.

Answer 14

\[\Large \begin{array}{l}
\frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\
{x^2} - 3x - 1 = 0\\
a = 1,b = - 3,c = - 1\\
\frac{{3 \pm \sqrt {{{( - 3)}^2} - (4 \cdot 1 \cdot - 1)} }}{{2 \cdot 1}}\\
= \frac{{3 \pm \sqrt {9 + 4} }}{2}\\
= \frac{{3 \pm \sqrt {13} }}{2}
\end{array}\]

Answer 15

Yep, you're right jemurray

Answer 16

Hero, how did we get this?

Answer 17

x^2-3x-1=0
x^2-3x= 1
x^2-3x + 9/4 = 1 + 9/4
(x-3/2)^2 = 13/4
x - 3/2 = +/-sqrt(13/4)
x - 3/2 = +/=sqrt(13)/2
x = +/-sqrt(13)/2 +3/2
x = (+/-sqrt(13) + 3)/2

Answer 18

4x^2 +x-1=0 will this be 8X^2-11

Answer 19

I think we should all show our methods. I don't think its good to tell someone there method is bad. Lets just all share and have fun. ;)

Answer 20

That's poor clarifying, myininaya.

Answer 21

ok so zonk you are asking how to solve 4x^2+x-1=0?

Answer 22

Sometimes telling someone their method is not that great is a good thing. If you're a teacher, you wouldn't lie to your student and say their method is good when it isn't. Total honesty applies here mya.

Answer 23

Well I wasn't using the quadratic formula meverett's method?

Answer 24

Thats the only method I see by him and I think it is a good one

Answer 25

yes

Answer 26

You come across as quite aggressive, condescending and belittling, Hero. If we're being honest. But this is straying further from the topic.

Answer 27

I agree with mya. The discussion initiated by hero in three forums today has been inappropriate, then he goes on and on and leaves the student hanging, never addressing the question. QED

Answer 28

\[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

Answer 29

Ok lets all just be cool okay?

Answer 30

meverett, you were the person whose methods I addressed, so of course you're not too happy with me.

Answer 31

I agree fsm and we are way off topic

I try to motivate not hate ...

Answer 32

I am not happy not because you disagree, but because being off topic is inappropriate and not helping the student.

Answer 33

lets help zonk with his problem

Answer 34

thanks

Answer 35

\[4x^2+x-1=0\]
so a=4,b=1,c=-1

Answer 36

we plugged this into the formula we have above

Answer 37

\[x=\frac{-(1) \pm \sqrt{(1)^2-4(4)(-1)}}{2(4)}\]

Answer 38

We could show this by completing the square.

Answer 39

How about you show him fool? The floor is yours.

Answer 40

\[x=\frac{-1 \pm \sqrt{1-16(-1)}}{8}=\frac{-1 \pm \sqrt{1+16}}{8}=\frac{-1 \pm \sqrt{17}}{8}\]

Answer 41

And I think fool wants to show you how to derive the quad formula

Answer 42

was that the end

Answer 43

lol

Answer 44

zonk would you like me to show you how the quad formula is derived?

Answer 45

How did this turn into a discussion about how to derive the quad formula. Seems like mya and fool are going at it head to head to me

Answer 46

Me encanta el chisme.

Answer 47

Let me derive in steps:

We want to solve the equation \(ax^2+bx+c=0\) where \( a \ne 0 \).

Answer 48

The usual argument starts by *dividing* by \(a \) let multiply both sides by \( 4a \)

Answer 49

We will get \( 4a^2x^2 +4abx+4ac=0 \)

Answer 50

\[ax^2+bx+c=0\]
a does not equal to zero
Divide both sides by a
\[x^2+\frac{b}{a}x+\frac{c}{a}=0\]
Subtract c/a on both sides
\[x^2+\frac{b}{a}x=-\frac{c}{a}\]
Now we add (b/2a)^2 on both sides (this completes the square on the left hand side
\[x^2+\frac{b}{a}x+(\frac{b}{2a})^2=\frac{-c}{a}+(\frac{b}{2a})^2\]
\[(x+\frac{b}{2a})^2=\frac{-c}{a}+\frac{b^2}{4a^2}\]
Now the goal is to solve for x
So we need to take square root of both sides
\[x+\frac{b}{2a}=\pm \sqrt{\frac{-c}{a}+\frac{b^2}{4a^2}}\]
Now subtract b/2a on both sides
\[x=-\frac{b}{2a} \pm \sqrt{\frac{-c}{a}+\frac{b^2}{4a^2}}\]
Now let' s combine the fractions inside the square root
\[x=\frac{-b}{2a} \pm \sqrt{\frac{-4ac+b^2}{4a^2}}\]
Now we can write
\[x=\frac{-b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a}\]
We can combine the fractions
\[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

Answer 51

OMG

Answer 52

okay now, \( 4a^2x^2+4abx \) is almost the square of \( 2ax+b \). Precisely, \( 4a^2x^2+4abx=(2ax+b)^2-b^2 \)

Answer 53

So, our equation reduces to \( (2ax+b)^2 -b^2+4ac=0 \Rightarrow (2ax+b)^2=b^2-4ac \)

Answer 54

it's almost done: \( 2ax+b=\pm\sqrt{b^2-4ac} \Rightarrow x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \)

Answer 55

Nice.

Answer 56

Notice there is no fraction till the end, that's the beauty of this approach!

Answer 57

the final question is
did you get it zonk ?

Answer 58

lol, he/she should by now.