GMAT DS : Quadratic Equation

A relatively easy Data Sufficiency question. Combines concepts in quadratic equation and inequalities involving quadratic expressions.

Is x = 3?
1. x² + 6x + 9 = 0
2. x² + 7x – 21 > 0

The correct answer is Choice A. Statement 1 alone is sufficient; statement 2 is not.

Explanatory Answer

The question to be answered is “Is x = 3?”.

An “Is” question is answered if you are able to provide a definite Yes or a definite No as an answer to the question. A “NO” is also a valid answer. Using the information in the statements if you are able to answer the question with a conclusive YES or a conclusive NO, the data is sufficient. Else it is not.

Statement 1
x² + 6x + 9 = 0

Factorizing, we get x² + 3x + 3x + 9 = 0

Or (x + 3)(x + 3) = 0.
So, we get x = -3.

Using the information in statement 1, we can conclusively answer the question x is NOT 3.
So, statement 1 is sufficient to answer the question.

Statement 2

x² + 7x – 21 > 0

A quick look at the quadratic expression – you will realize that one cannot factorize it.

There are two approaches at this stage. You could find the range of values that x takes using the formula to find the roots of the quadratic expression and then arrive at the answer.

The second approach is quicker.
Let us see if the inequality holds good when x = 3.

When x = 3, x² + 7x – 21 = 3² + 7(3) – 21 = 9 > 0

At x = 3, the inequality holds good.

Is 3 the only value for which this inequality will hold good or is there any other value for which the inequality will hold good?

Let us check with x = 4. When x = 4, x² + 7x – 21 = 4² + 7(4) – 21 = 23 > 0.

So, it is evident that the inequality holds good for 3 and many other values.
Therefore, we cannot conclude that x = 3 based on this inequality.

Hence, statement 2 is NOT sufficient.

Statement 1 is sufficient; statement 2 is not. Choice A is the correct answer

Let us look at few more variants to the two statements for the same question and understand how to crack these questions.

Variant 1

Is x = 3?
1. x² + 17x – 60 = 0
2. x – y² > 0

Statement 1 : x² + 17x – 60 = 0
Whenever, I look at a question such as “Is x = 3?” where the data is sufficient if I can arrive at a decisive Yes or a No, I will quickly check to see if substituting x as 3 in the equation satisfies it.

If it DOES NOT satisfy, I readily have an answer that says x is NOT 3. So, the data will be sufficient. Only if substituting x = 3 satisfies will I be tempted to find if any other value exists for x that will also satisfy this equation.

This approach saves a few seconds in avoiding the necessity to factorize a quadratic equation. More importantly, if the equation cannot be factorized as given in statement 2 of the original question, this method comes in handy.

So, let us get on with the process. Substitute x = 3 in the equation  x² + 17x – 60.
3² + 17(3) – 60 = 9 + 51 – 60 = 0. So,we know that x COULD be 3.

Now, we need to check if x could take a value other than 3. So, let us see if we can factorize this equation.
The equation factorizes as  x² + 20x – 3x + 60 = 0
or (x + 20)(x – 3) = 0.
So, x = -20 or x = 3.

The conclusion is that there is another value of x that satisfies this equation. So, x could either be 3 or it need not be 3.

Therefore, statement 1 is NOT SUFFICIENT.

Statement 2 :  x – y² > 0
x can take infinite values, one of which could be 3. So, statement 2 is NOT SUFFICIENT.

However, before discarding statement 2, let us take a quick look at what we can glean from statement 2.
We can rewrite the statement as x > y² .
For real values of y, y² will never be a negative number.

So, x which greater than y² has to necessarily be a positive number.

Combining Statement 1 and Statement 2 :  x² + 17x – 60 = 0 and  x – y² > 0

From statement 1, we know x takes two values : -20 or 3
From statement 2, we know that x has to be positive.

Using the information in the two statements together we can conclude that x has to be 3.

Choice C will the answer to this variant.

Variant 2

Is x = 3?
1. x² + 10x – 39 = 0
2. x + 10y² > 0

Statement 1 : x² + 10x – 39 = 0

As discussed earlier, let us quickly check if 3 could be a solution to the equation in statement 1.
When x = 3, the equation becomes 3² + 10(3) – 39 = 9 + 30 – 39 = 0.

So, 3 is one of the roots of this equation. x COULD be 3.

We have to factorize to check if any other value of x will satisfy the equation.
The equation factorizes as x² + 13x – 3x – 39 = 0
Or (x + 13)(x – 3) = 0
So, x = -13 or x = 3.

We can therefore, conclude that statement 1 is NOT SUFFICIENT.

Statement 2 :  x + 10y² > 0
Statement 2 alone is NOT SUFFICIENT as x could take infinite values, one of which could be 3.

Let us see what else can be gleaned from statement 2.
y² > 0 for real values of y.
So, 10y² > 0

Let us combine the two equation and evaluate the question

From statement 1, we know x can take two values -13 and 3. One of these values is negative and the other is positive.

Let us see if statement 2 helps us determine whether x is positive or negative.

10y² is a non negative number. Let us take an example and evaluate statement 2.
Let y = 2. So, 10y² = 40.

x + 10y² > 0.
If x = 3, one of the values that x can take (from statement 1), 3 + 40 = 43 > 0.
If x = -13, the other value that x can take (from statement 1), -13 + 40 = 27 > 0.

So despite combining the data in statement 1 and statement 2 together, we are not able to arrive an answer to whether x = 3.

Therefore, choice E is the correct answer.