Quadratics

[AlexanderA884]

Question:
Solve for x given that: (x^2 + 6x + 9) / (x+3) = 7

Why can't I factor (x^2+6x+9) / (x+3) to give me (x+3)^2/(x+3) and then cancel one set of the (x+3)'s to leave me with x+3 = 7 and thus x=4.

When I cross multiple (x+3) and 7 and then set the equation equal to zero I get two solutions x =4 and x==-3. I know this is correct but I don't understand why my first approach does not work.
Is my first approach ever appropriate? For example when solving for x if (x^2 -10z +25) = 9, I factor first and do not set equal to zero and get two solutions only because I take the square root of the left side of the equation. Why is this okay?

Thank You!

[Sage Pearce-Higgins]

Good question. The problems you mention are all associated with the special properties of zero. Let me give you a really simple equation: 10x = 3x You can manipulate this equation in a variety of ways (e.g. to show 7x = 0, and so x must be 0) but you cannot simply divide both sides by "x" - this would give you the nonsense statement that 10 = 3. What's the takeaway here? Be careful when dividing an equation by a variable - you have to account for the possibility that the variable is equal to zero. Note how GMAT often guards against this possibility in individual problems by stating that variables cannot equal zero.

Back to your first example: (x^2 + 6x + 9) / (x+3) = 7 Factorizing the top of the fraction is fine, but then cancelling (x+3) from both numerator and denominator is okay, only if (x+3) is an actual number, not zero. If in doubt, I suggest taking the safer method of cross multiplying and then factorizing the quadratic.

As for your second example, it's okay to square root an expression with variables, so long as you account for the positive and negative consequences. That's fine in this example - (x^2 -10z +25) = 9 - because you get +3 and -3 on the right hand side of your equation and thus the two solutions. Note that you could get just the same solutions if you subtracted 9 from both sides and factorized the equation as (x-8)(x-2) = 0

Note how the usual approach to solving quadratics relies on the special properties of zero. After all, if A multiplied by B equals zero then, logically, at least one of A or B has to equal zero. This doesn't apply to other numbers: e.g. if AxB = 10, then A and B have infinite solutions.

[JbhB682]

Hi Sage --

Why does the above not apply to this equation (or does it ?)

(Source : made up )

Equation : y^3 < y^2

Am i allowed to divide both sides by y^2 ?

My understanding is I CAN because y is not equal to zero

I came to the conclusion y is not equal to zero by plugging in zero in the equation

Given y = 0 does not work for above equation ..i think y has to be a number other than zero

[JbhB682]

Though now i am confused

-- if i were to see another equation --> y^4 > y ^3

If i divide by y^3 in this case, i will get y > 1

But with testing, actually the range is
--- y > 1 and
--- y < 0

[Sage Pearce-Higgins]

I think I can help here. First of all, to be pedantic, you're not dealing with equations, but with inequalities. You can apply most algebra to inequalities, but there's one really important difference. I'm sure you can see that 5 > 3 is a true inequality, no problem. However, the inequality -5 > -3 is false; it should be -5 < -3. What's going on here? When you multiply (or divide) an inequality by a negative number, you need to flip the signs. If you know that, e.g., t > 4, then -t < -4. Try out some cases to see this more clearly.

This creates a problem when we're trying to simplify inequalities with variables: we have to be very careful when dividing or multiplying by variables. In fact, unless we know whether the variable is positive or negative the simple message is: don't do it!

In your first example ( y^3 < y^2 ) it's fine to divide by y^2. As you said, y can't be zero, and a square number is always positive. The solution y < 1 is a good one.

In your second example ( y^4 > y ^3 ) it's not okay to divide by y^3, as this may be positive (if y is positive) or it may be negative (if y is negative). Danger! I would say that it's safer to work out the range of y by using logic and testing cases. If y is negative, then y^4 is positive, so this inequality works for all negative y. However, if y is a fraction between 0 and 1, then y^4 is going to be smaller than y^3. So this inequality is really saying y < 0 or y > 1.

By the way, well done for thinking about this stuff and checking rules by picking examples - it's a great way to get a deeper understanding of the concepts involved.

[JbhB682]

Question:
Solve for x given that: (x^2 + 6x + 9) / (x+3) = 7

Why can't I factor (x^2+6x+9) / (x+3) to give me (x+3)^2/(x+3) and then cancel one set of the (x+3)'s to leave me with x+3 = 7 and thus x=4.

When I cross multiple (x+3) and 7 and then set the equation equal to zero I get two solutions x =4 and x==-3. I know this is correct but I don't understand why my first approach does not work.

Hi Sage -- i was taking a look at this question and I have a follow up

If one were to cross multiply, you get two solutions [x = 4 and x = -3]

In fact, if i were to plug in x = 4 back into the equation [(x^2 + 6x + 9) / (x+3)], it gives me the number on the Right hand side = 7, thus indicating x = 4 is a viable solution

But if i were to actually plug in x = -3 back into the equation [(x^2 + 6x + 9) / (x+3)], i was hoping to see again the left hand side = right hand of the equation

Plugging in x = -3, I actually get [ 9 - 18 + 9 / -3 +3] or [0/0]

[0/0 is NOT equal to 7]

Doesn't plugging in x = -3 prove, x cannot equal -3 and in fact x = 4 is the ONLY solution to the equation

Maybe i am missing something here

[Sage Pearce-Higgins]

Good find - there's a problem here. When presented with an equation with a division bar, we need to make sure that the denominator of the division isn't zero as dividing by zero gives an undefined result. GMAT is very careful to observe this and often starts problems with a statement that some variable are not equal to zero, for example PS 90 in OG2020.

Now the issue is that we can't simply rule out the solution x=3 because of the division by zero. If x=3 then the left side of the equation is "undefined" and therefore the equation simply doesn't make sense. As far as I understand, we can neither rule out nor confirm that possibility.

We've now reached the edge of what you need to know for GMAT. They avoid this issue (with a small exception in DS 353 in OG2020).