# Homework Help: NEED understanding, binomial squares

1. Jan 23, 2009

### slmarais

1. The problem statement, all variables and given/known data

x^2 = 2/11x + 99/121

2. Relevant equations

3. The attempt at a solution

x^2 = 2/11 x + 99/121
x^2 - 2/11x - 99/121 = 0
x^2 - 2/11x =99/121

I understand that (b/2)^2 must be added to each side to become a perfect square trinomial....But HOW I do it is where Im stuck..I am suppper math challenged....please help!!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 23, 2009

### symbolipoint

Back-track by one step to: x^2 - 2/11x - 99/121 = 0, which you really should represent in some clearer way, since you really mean: x2 - $$\frac{2x}{11}$$ - $$\frac{99}{121}$$=0.

Next, multiply both sides by 121, because this is 112. This gives you a trinomial of the left side which may be easier to work with. You can either try factoring it, or use general solution to quadratic formula. Not sure if it is factorable; just try first to find out.

Last edited: Jan 23, 2009
3. Jan 23, 2009

### wsalem

Note that when solving the given equation $$x^2 - 2/11x - 99/121 = 0$$, you will not be able to represent it as a perfect square trinomial, since the definition of a perfect square trinomial $$x^2 \pm 2ax + a^2 = (x \pm a)^2$$ tell us that the sign of $$a^2$$ (which is $$\frac{-99}{121}$$ in your case) must be positive, but it is negative in the equation you're trying to solve.
So when you factor your equation you'll get an equation $$(x+a)(x-b)$$, i.e two values of x.

p.s If it is not obvious, $$\frac{99}{121} = \frac{9}{11}$$. So multiply your equation only by $$11$$, or you can just go ahead, choose the easiest!

4. Jan 23, 2009

### epenguin

Firstly pretty likely the middle term should have been written 2x/11 or (2/11)x.
Because that gives neat simple answer, otherwise you have got a cubic.

I would rather keep the last term as (9X11)/(11X11)

Can you think of any way involving 9/11 and `11/11 to somehow get -2/11??

5. Jan 23, 2009

### symbolipoint

Factoring did not help, so I tried directly using solution to quadratic equation. The result was complex. Anyone else try but find Real instead? Maybe my mistake?

The unrefined form I found was:

$$\frac{22\pm\sqrt{-3872}}{22}$$

6. Jan 23, 2009

### yeongil

No, the two solutions are both real (and rational, for that matter).
$$x^{2} = \frac{2}{11}x + \frac{99}{121}$$

$$x^{2} {-} \frac{2}{11}x = \frac{99}{121}$$

Yes, you need to find $$\left( \frac{b}{2}\right)^{2}$$. Since b is $$\frac{2}{11}$$, half of that is $$\frac{1}{11}$$, and squaring that result gives you $$\frac{1}{121}$$.

So, add that to both sides...
$$x^{2} {-} \frac{2}{11}x + \frac{1}{121} = \frac{99}{121} + \frac{1}{121}$$

$$x^{2} {-} \frac{2}{11}x + \frac{1}{121} = \frac{100}{121}$$

The left side is now a perfect square trinomial, and the right side is a perfect square. Can you take it from there?

01

7. Jan 23, 2009

### symbolipoint

The reported result in my post #5 in fact was based on a sign error so my answer was wrong. The original equation transforms to x^2 - (2/11)x - (99/121) = 0.

Clearing the fraction denominator transforms to 11x^2 - 22x - 99 = 0.

Resorting directly to the solution of a quadratic equation formula gives after simplifying,
x = 1 +/- (10)^(1/2)
OR
x = 1 $$\pm$$ $$\sqrt{10}$$

The solution is both Real and Irrational.

8. Jan 23, 2009

### slmarais

THANK YOU YEONGIL SOOOO MUCH!! I think Ive finally grasped it!!!
Thank you everyone!

x^2 = 2/11x + 1/121 = 100/121
(x - 1/11)² = 100/121
x - 1/11 = ± 10/11
x = (1 ± 10)/11
( 1, - 9/11)

9. Jan 24, 2009

### symbolipoint

Seemingly very good. I found another mistake in my work, this time (yielding post #7), was a multiplication mistake. I transformed into the wrong equation. Enough Done! Fine work, yeongil.

10. Jan 24, 2009

### epenguin

I believe the method I suggested in #4 of factorising

x2 - x.2/11 + 99/121

= x2 - x.2/11 + (9 X 11)/(11 X 11)

= (x - 11/11)(x + 9/11)

= (x - 1)(x + 9/11)

is called 'by inspection'.

smug