GMAT Math: Quadratic equation

Written by Kelly Granson. Posted in GMAT Study Guide

550px-Derive-the-Quadratic-Formula-Step-1Introduction

Quadratic equations, like any other arithmetic topic, have specific importance in GMAT. You can expect to see stand-alone quadratic equation problems and quadratic equation questions combined in problems with other topics such as number theory or progressions .

General form of a quadratic equation

Quadratic equations will be of the form 05.03, where a and b are coefficients of x and 05.031, and c is a constant term.

Solving a quadratic equation

A quadratic equation can be solved in two ways.

1) Factoring the middle term;

2) Using a quadratic formula 05.032.

Factoring the middle term is usually the first approach, but if you can't easily accomplish that, then go for the other method and simply use a formula to find the roots of the equation.

Example

Solve the expression 05.033.

First, try the factoring method for 05.033.

Step1. The middle term 5x has to be rewritten as two terms such that their product equals the product of the first and last terms : 05.033 becomes 05.034.

Step2. Take the GCF of the first two terms and the next two terms: x(x+2) + 3(x+2) = 0.

Step3. Take the GCF of the remaining terms: (x+2)*(x+3) = 0.

Step 4. Now equate each term to zero: x+ 2 = 0 and x+ 3 = 0. Hence, x = –2 or x = –3.

Problem

The quadratic expression 05.035 has two distinct roots. Find the number of possible values of p if the roots are integers.

A. 11

B. 12

C. 25

D. 19

E. 14

Solution Method A

For any quadratic equation of the form 05.03, the sum of the roots is –b/a and the product of the roots is c/a.

For the equation 05.035, the product of roots is c/a = 72/1 = 72.

Now find the number of values that p can take. If you can find all the possible combinations for the quadratic equation, then you can easily determine the possible values for p.

Since the roots must be integers, according to the problem statement...

The product of roots is 05.036, and both 05.037 and 05.038 are integers.

Now, to find the combinations of integers whose product is 72, start with 1: (1, 72), (2, 36), (3, 24), (4, 18), (6, 12) and (8, 9) where both the factors are positive, for a total of six combinations.

But the same multiples can also be two negatives: (-1, -72), (-2, -36), (-3, -24), (-4, -18), (-6, -12) and (-8, -9), for six more combinations.

Therefore, 12 combinations of integers are possible where the product of 05.039 and 05.0311 is 72.

Hence, p can take 24 possible values.

Solution Method B

If positive integer n has x integral factors, then it can be expressed as the product of two number is 05.0312 ways.

Find the number of possible factors for 72.

Write 72 as a product of its prime factors:

72 = 2* 36;

72= 2* 2* 18;

72 = 2* 2* 2* 9;

72 = 2* 2 * 2 * 3 * 3;

05.0313

So the number of factors is 05.0314, where 05.0315 and 05.0316 are exponents of prime numbers. Here the exponents are 3 and 2. Hence, the number of factors is (3 + 1)*(2 + 1) or 12 (increase the power of each prime factor by 1 and multiply the results).

Therefore, 72 has 12 positive integral factors and can be expressed as the product of any of six pairs of integers, each of which can be either two positive integers or two negative integers, for a total of twelve pairs of integers, or twenty-four possible values of p.

Option B is the correct answer.

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