Solve equation using quadratic formula

[David23454]

Homework Statement

I'm having trouble with solving a certain form of an equation with the quadratic formula. I think I'm making a dumb mistake somewhere with my algebra, but I can't seem to find it.

Solve with quadratic formula: 1/x^2=4/(0.02-x)^2

Homework Equations

1/x^2=4/(0.02-x)^2

The Attempt at a Solution

1/x^2=4/(0.02-x)^2
1/x^2=4/(0.0004-0.04x+x^2)
(1/4)(x^2)(1/x^2)=(1/4)(x^2)(4/(0.0004-0.04x+x^2))
1/4=x^2/(1/(0.0004-0.04x+x^2)
0.25=(x^2/0.0004)-(x^2/0.04x)+(x^2/x^2)
0.25=(2500x^2)-(25x)+1
0=(2500x^2)-(25x)+0.75

When I put this into the quadratic formula, I get an unreal solution, and I know I messed up on my algebra because when I solve the equation like this:

1/x^2=4/(0.02-x)^2 (Cross multiply)
0.0004-0.04x+x^2=4x^2
3x^2+0.04x-0.0004=0

...I get the correct answer, which is (0.00667, -0.02). Could someone show me where I made my mistake? Thanks!

 

[andrewkirk]

In the fourth line you have removed the factor (1/4) from the right side but not from the left.

 

[Chestermiller]

Was it specified that you had to solve this problem using the quadratic formula?

 

[David23454]

Sorry! I copied it down wrong--with the four canceled it does give the equation 0=(2500x^2)-(25x)+0.75 which doesn't work in the quadratic formula.

1/x^2=4/(0.02-x)^2
1/x^2=4/(0.0004-0.04x+x^2)
(1/4)(x^2)(1/x^2)=(1/4)(x^2)(4/(0.0004-0.04x+x^2))
1/4=x^2/(1/(0.0004-0.04x+x^2)
0.25=(x^2/0.0004)-(x^2/0.04x)+(x^2/x^2)
0.25=(2500x^2)-(25x)+1
0=(2500x^2)-(25x)+0.75

Any idea where I went wrong?

 

[David23454]

No, but I wanted to see what I did wrong with the equation to find out what I did wrong with my algebra so I don't make the same mistake again later.

 

[Chestermiller]

No, but I wanted to see what I did wrong with the quadratic equation solution, to find out what I did wrong with my algebra so I don't make the same mistake again later.

If I take the reciprocal of both sides of the equation, I get:
$$x^2=\frac{(0.02-x)^2}{4}$$What do you get if you take the square root of both sides of this equation?

 

[David23454]

If I take the reciprocal of both sides of the equation, I get:
$$x^2=\frac{(0.02-x)^2}{4}$$What do you get if you take the square root of both sides of this equation?

x=0.00667, which is correct. I had figured out the correct answer already, but I was just curious about what I did wrong in the other method, just so I could know where my algebra knowledge was lacking, to avoid future mistakes.

 

[Chestermiller]

$$x^2=\frac{(0.02-x)^2}{4}$$
Taking the square root of both sides, one obtains:$$x=\pm \frac{(0.02-x)}{2}$$or equivalently:
$$x=+ \frac{(0.02-x)}{2}\tag{1}$$
and
$$x=- \frac{(0.02-x)}{2}\tag{2}$$Eqns. 1 and 2 represent very simple linear algebraic equations for x. The solution to Eqn. 1 is $x=0.00667$. The solution to Eqn. 2 is $x=-0.02$.

The key learning from all this is that sometimes it is worthwhile to look for a simple "trick" to solve a problem, rather than always trying to solve problems by brute force.

 

[David23454]

The key learning from all this is that sometimes it is worthwhile to look for a simple "trick" to solve a problem, rather than always trying to solve problems by brute force.

I understand this, but as I mentioned in the first post, I already had found the correct solution, but I wanted to know where I went wrong initially with my algebra in the process I used, so I could pinpoint what I had done wrong to avoid doing so again in the future.

 

[Chestermiller]

I understand this, but as I mentioned in the first post, I already had found the correct solution, but I wanted to know where I went wrong initially with my algebra in the process I used, so I could pinpoint what I had done wrong to avoid doing so again in the future.

I realize that, but my point was that it often makes life much easier on you if you look for simple tricks to solve problems. This concept does not just apply to this particular problem. My objective was to change your way of thinking about how to approach problems. The present problem is merely incidental to this concept.

 

[David23454]

Ok, I understand your point, and I'll certainly use the method you demonstrated to solve similar problems in the future, but I'm still very curious as to what I did wrong in my attempt.

 

[Chestermiller]

Ok, I understand your point, and I'll certainly use the method you demonstrated to solve similar problems in the future, but I'm still very curious as to what I did wrong in my attempt.

You want someone to check your algebra, correct?

 

[David23454]

yes

 

[Chestermiller]

Ok, I understand your point, and I'll certainly use the method you demonstrated to solve similar problems in the future, but I'm still very curious as to what I did wrong in my attempt.

I wasn't only referring to "similar problems." I was referring to all math problems you encounter.

 

[Chestermiller]

yes

I'll let someone else help you with that. It goes against every fiber of my being to solve a problem by brute force.

 

[andrewkirk]

Sorry! I copied it down wrong--with the four canceled it does give the equation 0=(2500x^2)-(25x)+0.75 which doesn't work in the quadratic formula.

1/x^2=4/(0.02-x)^2
1/x^2=4/(0.0004-0.04x+x^2)
(1/4)(x^2)(1/x^2)=(1/4)(x^2)(4/(0.0004-0.04x+x^2))
1/4=x^2 / (1/(0.0004-0.04x+x^2)

that red slash should not be there.

The line following that is also wrong. 1/(a+b+c) is not equal to 1/a + 1/b + 1/c.