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Simple Quadratic Equation

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  • #1
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Homework Statement


Find the exact (surd) answer


Homework Equations



-2kx^2 + 4x +6k = 0


The Attempt at a Solution



Attempt 1 (quadratic formula method)

-2kx^2 + 4x +6k = 0

-> -(-2) +- sqrt((-2)^2 - (4 x k x -3k)0 / 2k

-> 2 +- sqrt(4+12k^2) / 2k

-> 2 +- sqrt(4(1+3k^2) / 2k

-> 2 +- sqrt(4) sqrt(1+3k^2) / 2k

-> 2 +- 2sqrt(1+3k^2) / 2k

-> x = sqrt(1 + 3k^2) / k

Given solutions are:

x = sqrt(3k^2 + 1 + 1) / k

and

x = -sqrt(3k^2 + 1 - 1) / k


Attempt 2 (completing the square method)

-2kx^2 + 4x + 6k = 0

-> kx^2 - 2x - 3k = 0

-> kx^2 - 2x + 1 -1 - 3k = 0

-> kx^2 - 2x + 1 = 1 + 3k

Not sure how to factor kx^2 - 2x + 1 but if I move k to the RHS, wouldn't the k's cancel out? I'm interested in solving it both ways as I'm trying to teach myself this and I don't want any holes in my knowledge

Thanks a lot
 

Answers and Replies

  • #2
33,506
5,191

Homework Statement


Find the exact (surd) answer


Homework Equations



-2kx^2 + 4x +6k = 0
Do some simplification first by dividing both sides by -2, to get
kx^2 - 2k - 3k = 0

The Attempt at a Solution



Attempt 1 (quadratic formula method)

-2kx^2 + 4x +6k = 0

-> -(-2) +- sqrt((-2)^2 - (4 x k x -3k)0 / 2k
Not sure what you're doing above. Make sure you use enough parentheses. The final 2k is under everything.
-> 2 +- sqrt(4+12k^2) / 2k
Should be
x = [2 +/- sqrt(4 + 12k^2)]/(2k)

Each step here should be an equation, or at least indicate that each expression is equal to the one above.
-> 2 +- sqrt(4(1+3k^2) / 2k

-> 2 +- sqrt(4) sqrt(1+3k^2) / 2k

-> 2 +- 2sqrt(1+3k^2) / 2k

-> x = sqrt(1 + 3k^2) / k
No, the numerator should still have two terms.

x = [1 +/- sqrt(1 + 3k^2)]/k
Given solutions are:

x = sqrt(3k^2 + 1 + 1) / k

and

x = -sqrt(3k^2 + 1 - 1) / k
Neither of the above is correct. Are you sure you're writing them correctly?
Attempt 2 (completing the square method)

-2kx^2 + 4x + 6k = 0

-> kx^2 - 2x - 3k = 0

-> kx^2 - 2x + 1 -1 - 3k = 0

-> kx^2 - 2x + 1 = 1 + 3k

Not sure how to factor kx^2 - 2x + 1 but if I move k to the RHS, wouldn't the k's cancel out? I'm interested in solving it both ways as I'm trying to teach myself this and I don't want any holes in my knowledge

Thanks a lot
 
  • #3
333
0
No, the numerator should still have two terms.

x = [1 +/- sqrt(1 + 3k^2)]/k
Sorry but I don't understand where the 1 comes from...if I cancel the twos, shouldn't I just be left with x = [+/- sqrt(1 + 3k^2)]/k


Yeah the second given solution is definitely written out like that, or maybe like this

x = [sqrt(3k^2 + 1 + 1)] / k

and

x = [-sqrt(3k^2 + 1 - 1)] / k
 
  • #4
901
2
Find the value of k by using the discriminant [tex]\Delta = b^2 - 4ac[/tex].
 
  • #5
33,506
5,191
Sorry but I don't understand where the 1 comes from...if I cancel the twos, shouldn't I just be left with x = [+/- sqrt(1 + 3k^2)]/k
This is a very common mistake.

This is something like saying that
[tex]\frac{2 + 6}{2} = 6[/tex]

I hope you recognize that the actual answer is 4.

The only time you can cancel is when the numerator and denominator are factored, which is not the case in my example or in your problem.

[tex]x = \frac{2 \pm 2\sqrt{1 + 3k^2}}{2k} = \frac{2(1 \pm \sqrt{1 + 3k^2})}{2k}[/tex]

Now 2 is a factor in both numerator and denominator, so now we can cancel.

[tex]x = \frac{1 \pm \sqrt{1 + 3k^2}}{k}[/tex]

Yeah the second given solution is definitely written out like that, or maybe like this

x = [sqrt(3k^2 + 1 + 1)] / k

and

x = [-sqrt(3k^2 + 1 - 1)] / k
I have no idea where they got these. Both look like they have typos. Why would anyone write 1 + 1 and 1 - 1 in a final answer?
 
  • #6
333
0
^ Ah it makes sense now...cheers!
 
  • #7
33,506
5,191
Find the value of k by using the discriminant [tex]\Delta = b^2 - 4ac[/tex].
It is not possible to find the value of k in this problem.
 
  • #8
43
0
x = [sqrt(3k^2 + 1 + 1)] / k

and

x = [-sqrt(3k^2 + 1 - 1)] / k
I have no idea where they got these. Both look like they have typos. Why would anyone write 1 + 1 and 1 - 1 in a final answer?
The brackets are written incorrectly. Instead of writing:

x = [sqrt(3k^2 + 1 + 1)] / k
x = [-sqrt(3k^2 + 1 - 1)] / k

It should be:

x = [sqrt(3k^2+1) + 1] /k
x = {-[sqrt(3k^2+1) - 1]} /k

which is truly confusing, I have no clue why they would decide to put the 1 after the square root sign.

It's really just another way to write:

x = [1 +/- sqrt(3k^2+1)] / k
 
  • #9
43
0
It's confusing with all those brackets, but what I did is analogous to the following:

a - b = -(b - a)

For example, with 5 representing what's before the +/- sign (for the negative case only) and 4 representing what's under the square root sign:

5 - 4 = -(4 - 5)
1 = -4 + 5
1 = 1

-(b+a) doesn't work:
= -(5 + 4)
= -5 - 4
= -9

At least that's my reasoning via the distributive property.
 
  • #10
33,506
5,191
It's confusing with all those brackets, but what I did is analogous to the following:

a - b = -(b - a)
OK, I see what you did. I didn't notice that you had inserted another pair of brackets.
For example, with 5 representing what's before the +/- sign (for the negative case only) and 4 representing what's under the square root sign:

5 - 4 = -(4 - 5)
1 = -4 + 5
1 = 1

-(b+a) doesn't work:
= -(5 + 4)
= -5 - 4
= -9

At least that's my reasoning via the distributive property.
 
  • #11
333
0
Yeah, that must be it :p

How would you solve it by completing the square?
 
  • #12
HallsofIvy
Science Advisor
Homework Helper
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When you have an equation of the form [itex]ax^2+ bx+ c= 0[/itex] with [itex]a\ne 1[/itex], the simplest way to use "completing the square" is to first divide through by a.

You have [itex]-2kx^2 + 4x +6k = 0[/itex] so divide through by -2k:
[tex]x^2- \frac{2}{k}x- 3= 0[/tex]
and then write it as
[tex]x^2- \frac{2}{k}x= 3[/tex]

Now compare that to [itex](x- a)^2= x^2- 2ax+ a^2[/itex]. with
[tex]-\frac{2}{k}x= -2ax[/tex]
what is a? What do you need to add to both sides to get a "perfect square" on the left?
 
  • #13
43
0
I just though I would mention something you might want to try. If you take [itex]ax^2+ bx+ c= 0[/itex] and factor it the way that HallsofIvy says, you'll actually end up with the quadratic formula.

Just in case you wanted to see where the quadratic formula comes from.

Edit: This video might help you understand the process if you get stuck.

https://www.youtube.com/watch?v=mDmRYfma9C0
 
Last edited:
  • #14
333
0
When you have an equation of the form [itex]ax^2+ bx+ c= 0[/itex] with [itex]a\ne 1[/itex], the simplest way to use "completing the square" is to first divide through by a.

You have [itex]-2kx^2 + 4x +6k = 0[/itex] so divide through by -2k:
[tex]x^2- \frac{2}{k}x- 3= 0[/tex]
and then write it as
[tex]x^2- \frac{2}{k}x= 3[/tex]

Now compare that to [itex](x- a)^2= x^2- 2ax+ a^2[/itex]. with
[tex]-\frac{2}{k}x= -2ax[/tex]
what is a? What do you need to add to both sides to get a "perfect square" on the left?
Don't you have to add and subtract (b/2)^2 as well. So -2x/k is the same thing as -(2/k)x?
Sorry there are lot of holes in my math knowledge.

@ebits21: thanks I'll check it out
 

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