Quadratic Equations Notes and Tricks for Banking and Competitive

Quadratic equations are a favourite topic of the preliminary and mains examinations of banking (SBI, IBPS), insurance and other competitive tests. In the banking exams, the questions are mostly of one type: two equations are given, you find the values of x and y, and then you compare them. The questions are not difficult, but they need speed and a clear method. This note explains what a quadratic equation is, three ways to solve it, the shortcuts for the sign of the roots, the standard comparison type with a method to answer in 30 seconds, and word problems, all in simple language with solved examples.

What Is a Quadratic Equation?

A quadratic equation is an equation of the form ax squared + bx + c = 0, where a, b and c are numbers and a is not zero. The highest power of x is 2, so the equation has two roots (two values of x that satisfy it). For example, x squared - 7x + 12 = 0 has the roots 3 and 4, because each of them makes the left side zero.

Three Methods to Solve

1. Splitting the Middle Term (Factorisation)

This is the fastest method for the equations in banking exams. Split the middle term b into two parts whose product is a x c and whose sum is b. Example: x squared - 7x + 12 = 0. Here a x c = 12, and the numbers whose product is 12 and the sum is -7 are -3 and -4. So the equation becomes x squared - 3x - 4x + 12 = 0, that is x(x - 3) - 4(x - 3) = 0, or (x - 3)(x - 4) = 0. The roots are x = 3 and x = 4.

2. The Quadratic Formula

If the equation does not factorise easily, use x = [-b plus or minus the square root of (b squared - 4ac)] / 2a. The quantity D = b squared - 4ac is called the discriminant. Example: x squared - 4x - 5 = 0. D = 16 + 20 = 36, and the square root is 6. So x = (4 + 6)/2 = 5 or x = (4 - 6)/2 = -1.

3. Sum and Product of Roots

For ax squared + bx + c = 0, the sum of the roots = -b/a and the product of the roots = c/a. This is useful to check your answer. For 2x squared - 8x + 6 = 0, the sum is 8/2 = 4 and the product is 6/2 = 3, which agrees with the roots 1 and 3.

Nature of the Roots

Shortcut: Signs of the Roots

You can guess the signs of the roots before solving, from the signs of b and c (taking a as positive):

For example, x squared + 5x + 6 = 0 has both c and b positive, so both roots are negative; they are -2 and -3. In x squared - 5x + 6 = 0, c is positive and b is negative, so the roots are positive, namely 2 and 3.

The Banking Question Type: Compare x and y

A typical question reads: "I. x squared - 7x + 12 = 0; II. y squared - 5y + 6 = 0. Give the answer: (a) if x is greater than y, (b) if x is smaller than y, (c) if x is greater than or equal to y, (d) if x is smaller than or equal to y, (e) if x equals y or no relation can be established." The method is as follows.

1. Solve each equation by splitting the middle term and write both roots.
2. Put the roots of x and y on a number line, or compare each root of x with each root of y.
3. If every root of x is greater than every root of y, then x is greater than y. If every root of x is greater than or equal to every root of y, with some equal, then x is greater than or equal to y. The same logic applies to the less than cases.
4. If some comparisons give "greater" and some give "smaller", then there is no relation.

Solved Examples

1. I. x squared - 7x + 12 = 0; II. y squared - 5y + 6 = 0. x = 3, 4 and y = 2, 3. The comparisons are 3 is greater than 2, 3 equals 3, 4 is greater than 2 and 4 is greater than 3. Every x is at least as large as every y, so x is greater than or equal to y.
2. I. x squared + 5x + 6 = 0; II. y squared + 7y + 12 = 0. x = -2, -3 and y = -3, -4. Here -2 is greater than -3 and -4; -3 equals -3 and is greater than -4. So x is greater than or equal to y.
3. I. x squared - 9 = 0; II. y squared - 4y + 3 = 0. x = 3, -3 and y = 1, 3. Since 3 is greater than 1, but -3 is smaller than 1, no relation can be established.
4. I. 2x squared - 7x + 3 = 0; II. y squared - 3y + 2 = 0. The first gives (2x - 1)(x - 3) = 0, so x = 1/2 or 3. The second gives y = 1 or 2. Since 1/2 is smaller than 1, but 3 is greater than 2, no relation.

Word Problems on Quadratic Equations

• Sum and its square: The sum of a number and its square is 42. Find the number. Let the number be x. x squared + x - 42 = 0, which gives (x + 7)(x - 6) = 0, so the number is 6 or -7.
• Area problem: The length of a rectangle is 3 m more than its breadth, and the area is 70 square m. Find the breadth. Let the breadth be b: b(b + 3) = 70, so b squared + 3b - 70 = 0, which gives (b + 10)(b - 7) = 0. The breadth cannot be negative, so it is 7 m and the length is 10 m.
• Consecutive numbers: The product of two consecutive positive integers is 56. They are 7 and 8, since x(x + 1) = 56 gives x squared + x - 56 = 0, or (x + 8)(x - 7) = 0.

Tips to Solve Faster

• Learn the factor pairs of numbers up to 100, so that you can split the middle term quickly.
• Check the signs of the roots first using the sign table; this prevents the sign errors.
• Keep both roots of x and y written clearly in a line and compare them carefully.
• If the first term has a coefficient other than 1, multiply a and c to find the pair of numbers.
• Do not spend more than 40 seconds on a pair of equations; if it does not factorise, use the formula.

Common Mistakes

Many students make errors with the sign of the roots: they solve x - 3 = 0 correctly but write the root as -3. Others stop after finding one root. In the comparison type, the common error is to compare only one root of each equation and choose a wrong answer. Remember that you have to check all four pairs of comparisons. Also, do not confuse the "x is greater than y" option with "x is greater than or equal to y", as the equal case decides between the two.

Practice Questions With Answers

• Q1. Solve x squared - 5x + 6 = 0. Answer: x = 2 and 3.
• Q2. Solve x squared + x - 12 = 0. Answer: x = 3 and -4.
• Q3. Find the sum and the product of the roots of 2x squared - 8x + 6 = 0. Answer: Sum 4; product 3.
• Q4. Find the nature of the roots of x squared - 6x + 9 = 0. Answer: Real and equal (D = 0); the root is 3.
• Q5. I. x squared - 4 = 0; II. y squared - 5y + 6 = 0. Compare x and y. Answer: x = 2, -2 and y = 2, 3. Every x is less than or equal to every y, so x is less than or equal to y.

Read it together with our notes on algebra and the inequalities chapter of reasoning.

Quick Revision

• ax squared + bx + c = 0; roots by splitting the middle term or the formula.
• D = b squared - 4ac; sum = -b/a; product = c/a.
• Signs: c positive and b positive gives both negative roots; c positive and b negative gives both positive.
• For the comparison type, compare all the roots of x with all of y; mixed results mean no relation.

Frequently Asked Questions (FAQ)

What is a quadratic equation?

A quadratic equation is an equation of the form ax squared + bx + c = 0, where the highest power of the variable is 2.

How many roots does a quadratic equation have?

A quadratic equation has two roots.

What is the quadratic formula?

x = [-b plus or minus the square root of (b squared - 4ac)] divided by 2a.

What is the discriminant?

The discriminant is b squared - 4ac, and it tells the nature of the roots.

What is the sum of the roots of a quadratic equation?

The sum of the roots equals -b divided by a.

What is the product of the roots?

The product of the roots equals c divided by a.

What does D less than 0 mean?

It means that the equation has no real roots.

How can I find the signs of the roots quickly?

If c and b are both positive, both roots are negative; if c is positive and b is negative, both roots are positive; if c is negative, the roots have opposite signs.

How do I solve the x and y comparison questions in banking exams?

Solve both equations, write all the roots, compare each root of x with each root of y and decide; if results differ, there is no relation.

What is the method of splitting the middle term?

Split b into two numbers whose product is a times c and whose sum is b, and then factorise.

When should I use the quadratic formula?

Use it when the equation does not factorise easily.

How much time should I spend on a pair of quadratic equations?

About 40 seconds to a minute in the examination.