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CrackVerbal Representative G
Affiliations: CrackVerbal
Joined: 03 Oct 2013
Posts: 2013
Location: India
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Quadratic inequalities are an important and often overlooked concept on the GMAT. Once mastered, this concept can be used for any inequality involving polynomials and makes solving a complex inequality question, a mere walk in the park.

Before we start, let us recall that the standard form of a quadratic equation is ax^2 + bx + c = 0. The standard quadratic equation becomes an inequality if it is represented as ax^2 + bx + c < 0 or (>0).

Steps to Solve a Quadratic Inequality

Let us consider the quadratic inequality x^2 – 5x < -6.

1. Maintain the quadratic inequality in its standard form ax^2 + bx + c < 0 or (>0). If the inequality is not in the standard form then rewrite the inequality so that all nonzero terms appear on one side of the inequality sign.

Adding 6 on both sided of the above inequality we get x^2 – 5x + 6 < 0.

2. Factor the nonzero side of the inequality

x^2 – 5x + 6 < 0 ----> (x-2) (x-3) < 0
2 and 3 are the factors of the inequality.

3. Place the factors on number line. The number line will get divided into the three regions. Start from the right and mark the region with + sign, the next region with a – sign and the third region with a + sign (alternating + and - starting from the right).

Why the alternating + and – signs from the right hand side you may ask?

(x-2) (x-3) < 0. Placing factors 2 and 3 on the number line

Attachment: Capture.PNG [ 883 Bytes | Viewed 10168 times ]

If we consider any value to the right of 3 then the product (x-2) (x-3) will always be positive.
If we consider any value in between 2 and 3 then the product (x-2) (x-3) will always be negative.
If we consider any value to the left of 2 then the product (x-2) (x-3) will always be positive.

4. If the inequality is of the form ax^2 + bx + c < 0, the region having the - sign will be the solution of the given quadratic inequality. If the inequality is of the form ax^2 + bx + c > 0 the region having the + sign will be the solutions of the given quadratic inequality.

Since the inequality here is (x-2) (x-3) < 0, we need to consider the region on the number line having the negative sign. So the solution to the quadratic inequality x^2 – 5x + 6 < 0 is 2 < x < 3.

Other Applications

1. This procedure can be used to solve any inequality involving polynomials (not just quadratics).

Consider the cubic inequality (x+2)(x-2)(x-3) > 0 with factors -2, 2 and 3. Plotting the factors on the number line and alternating + and – starting from the right

Attachment: Capture 1.PNG [ 992 Bytes | Viewed 10169 times ]

Now since the inequality (x+2)(x-2)(x-3) > 0, we need to consider the positive regions on the number line. The solutions here are -2 < x < 2 and x > 3.

2. This procedure can also be used to solve algebraic inequality expressions which are in the form of fractions.

Consider the algebraic expression (x – 2) /(x – 3) ≥ 0. For the inequality to hold true both the numerator (x – 2) and the denominator (x – 3) have to be positive (case 1) or both negative (case 2). We can transform the expression (x – 2) /(x – 3) ≥ 0 to (x – 2) (x – 3) ≥ 0 since the product of (x – 2) and (x – 3) will also require both terms to be either positive or negative. Now this transformed expression is a quadratic inequality. The solution to the inequality (x – 2) (x – 3) ≥ 0 using the above discussed procedure is x ≥ 3 and x ≤ 2.

The only extra step we need to apply while solving the transformed equation is that we ignore the solution x = 3, since the original question (x – 2) /(x – 3) ≥ 0 has the term x – 3 in the denominator and the denominator of a fraction can never be 0. So the exact solution to the inequality (x – 2) /(x – 3) ≥ 0 is x > 3 and x ≤ 2.

Putting it all together

With these procedures in mind, let us try to solve a hard problem from the Official Guide. The reason I am choosing this particular problem is because the OG explanation is too confusing and convoluted and runs to almost a page! The Official guide is a great resource for questions but a horrible source of explanations. Especially for math, OG explanations are rarely the most efficient way to solve problems.

How many of the integers that satisfy the inequality ((x+2) (x+3))/(x-2) ≥ 0 are less than 5?
A. 1
B. 2
C. 3
D. 4
E. 5

The inequality ((x+2) (x+3))/(x-2) ≥ 0 can be transformed into (x+2) (x+3) (x-2) ≥ 0. Plotting the factors on the number line and alternating + and – signs starting from the right we get

Attachment: Capture 3.PNG [ 1008 Bytes | Viewed 10166 times ]

Since the inequality is (x+2) (x+3) (x-2) ≥ 0 we consider only the positive regions of the number line. The solution here is -3 ≤ x ≤ -2 and x ≥ 2. Here we need to ignore the value of x = 2 since this would make the denominator 0. The exact solution for the inequality ((x+2) (x+3))/(x-2) ≥ 0 is -3 ≤ x ≤ -2 and x > 2. The integer values less than 5 that fall under the two solution ranges are -3, -2, 3 and 4. The answer here is D.

Hope these questions don’t look very intimidating now. Use this method for your inequality problems; they won’t feel like problems anymore!

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Cheers!
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Originally posted by CrackVerbalGMAT on 27 Dec 2016, 13:27.
Last edited by CrackVerbalGMAT on 13 Dec 2017, 00:26, edited 2 times in total.
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CrackVerbalGMAT wrote:
Quadratic inequalities are an important and often overlooked concept on the GMAT. Once mastered, this concept can be used for any inequality involving polynomials and makes solving a complex inequality question, a mere walk in the park.

Before we start, let us recall that the standard form of a quadratic equation is ax^2 + bx + c = 0. The standard quadratic equation becomes an inequality if it is represented as ax^2 + bx + c < 0 or (>0).

Steps to Solve a Quadratic Inequality

Let us consider the quadratic inequality x^2 – 5x < -6.

1. Maintain the quadratic inequality in its standard form ax^2 + bx + c < 0 or (>0). If the inequality is not in the standard form then rewrite the inequality so that all nonzero terms appear on one side of the inequality sign.

Adding 6 on both sided of the above inequality we get x^2 – 5x + 6 < 0.

2. Factor the nonzero side of the inequality

x^2 – 5x + 6 < 0 ----> (x-2) (x-3) < 0
2 and 3 are the factors of the inequality.

3. Place the factors on number line. The number line will get divided into the three regions. Start from the right and mark the region with + sign, the next region with a – sign and the third region with a + sign (alternating + and - starting from the right).

Why the alternating + and – signs from the right hand side you may ask?

(x-2) (x-3) < 0. Placing factors 2 and 3 on the number line

Attachment:
Capture.PNG

If we consider any value to the right of 3 then the product (x-2) (x-3) will always be positive.
If we consider any value in between 2 and 3 then the product (x-2) (x-3) will always be negative.
If we consider any value to the left of 2 then the product (x-2) (x-3) will always be positive.

4. If the inequality is of the form ax^2 + bx + c < 0, the region having the - sign will be the solution of the given quadratic inequality. If the inequality is of the form ax^2 + bx + c > 0 the region having the + sign will be the solutions of the given quadratic inequality.

Since the inequality here is (x-2) (x-3) < 0, we need to consider the region on the number line having the negative sign. So the solution to the quadratic inequality x^2 – 5x + 6 < 0 is 2 < x < 3.

Other Applications

1. This procedure can be used to solve any inequality involving polynomials (not just quadratics).

Consider the cubic inequality (x+2)(x-2)(x-3) > 0 with factors -2, 2 and 3. Plotting the factors on the number line and alternating + and – starting from the right

Attachment:
Capture 1.PNG

Now since the inequality (x+2)(x-2)(x-3) > 0, we need to consider the positive regions on the number line. The solutions here are -2 < x < 2 and x > 3.

2. This procedure can also be used to solve algebraic inequality expressions which are in the form of fractions.

Consider the algebraic expression (x – 2) /(x – 3) ≥ 0. For the inequality to hold true both the numerator (x – 2) and the denominator (x – 3) have to be positive (case 1) or both negative (case 2). We can transform the expression (x – 2) /(x – 3) ≥ 0 to (x – 2) (x – 3) ≥ 0 since the product of (x – 2) and (x – 3) will also require both terms to be either positive or negative. Now this transformed expression is a quadratic inequality. The solution to the inequality (x – 2) (x – 3) ≥ 0 using the above discussed procedure is x ≥ 3 and x ≤ 2.

The only extra step we need to apply while solving the transformed equation is that we ignore the solution x = 3, since the original question (x – 2) /(x – 3) ≥ 0 has the term x – 3 in the denominator and the denominator of a fraction can never be 0. So the exact solution to the inequality (x – 2) /(x – 3) ≥ 0 is x > 3 and x ≤ 2.

Putting it all together

With these procedures in mind, let us try to solve a hard problem from the Official Guide. The reason I am choosing this particular problem is because the OG explanation is too confusing and convoluted and runs to almost a page! The Official guide is a great resource for questions but a horrible source of explanations. Especially for math, OG explanations are rarely the most efficient way to solve problems.

How many of the integers that satisfy the inequality ((x+2) (x+3))/(x-2) ≥ 0 are less than 5?
A. 1
B. 2
C. 3
D. 4
E. 5

The inequality ((x+2) (x+3))/(x-2) ≥ 0 can be transformed into (x+2) (x+3) (x-2) ≥ 0. Plotting the factors on the number line and alternating + and – signs starting from the right we get

Attachment:
Capture 3.PNG

Since the inequality is (x+2) (x+3) (x-2) ≥ 0 we consider only the positive regions of the number line. The solution here is -3 ≤ x ≤ -2 and x ≥ 2. Here we need to ignore the value of x = 2 since this would make the denominator 0. The exact solution for the inequality ((x+2) (x+3))/(x-2) ≥ 0 is -3 ≤ x ≤ -2 and x > 2. The integer values less than 5 that fall under the two solution ranges are -3, -2, 3 and 4. The answer here is D.

Hope these questions don’t look very intimidating now. Use this method for your inequality problems; they won’t feel like problems anymore!

Cheers!

Hello CrackVerbal, I have the below doubt

after finding the roots of the equation and putting then on a line - How do we decide whether we start from "+" or "-" ,,,, in two of your example one line starts from "+" and one line starts with "-"

Kindly explain.
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sriamlan wrote:

Hello CrackVerbal, I have the below doubt

after finding the roots of the equation and putting then on a line - How do we decide whether we start from "+" or "-" ,,,, in two of your example one line starts from "+" and one line starts with "-"

Kindly explain.

Hi,

We always start from the right and take that as +ve. Then we move to the left, taking alternate signs.

I mean right most will be +ve, then the one to its left will be -ve, then +ve , then -ve and so on.

Let me know in case it is not clear. _________________
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