{"id":5402665,"date":"2020-07-06T09:10:06","date_gmt":"2020-07-06T03:40:06","guid":{"rendered":"https:\/\/entri.app\/blog\/?p=5402665"},"modified":"2023-11-14T10:04:48","modified_gmt":"2023-11-14T04:34:48","slug":"tips-and-tricks-to-solve-quadratic-equation-for-bank-exam","status":"publish","type":"post","link":"https:\/\/entri.app\/blog\/tips-and-tricks-to-solve-quadratic-equation-for-bank-exam\/","title":{"rendered":"Tips and Tricks to Solve Quadratic Equation in Bank Exams"},"content":{"rendered":"

The Quadratic Equation is an important topic in the Quantitative Aptitude section. Questions from this topic are asked in almost every competitive exam. Hence, let’s check out some Tips and Tricks to solve Quadratic Equation Bank Exams here. Before you start solving the Quadratic Equation questions, you must know the definition of it. What actually is the Quadratic Equation.<\/span><\/p>\n

So, a Quadratic Equation is an equation that contains at least one squared term. It can be defined in the form of ax<\/span>2<\/span> + bx + c = 0. Where a, b, and c are constants and \u2018x\u2019 is an unknown variable.<\/span><\/p>\n

To clear the concept of the Quadratic Equation, you must be familiar with all the terminologies that are used.<\/span><\/p>\n

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\"Banking-2.0_rectangle_bannner\"<\/a><\/p>\n

Quadratic Expression<\/b><\/h3>\n

\u00a0<\/span>A Quadratic Expression is something that can be simplified or factored. An expression cannot be solved for a variable but you can manipulate the terms that are given. Examples of a Quadratic Expression are:<\/span><\/p>\n

x<\/span>2<\/span> + 2x + 2<\/span><\/p>\n

8x<\/span>2<\/span> \u2013 2x<\/span><\/p>\n

4 \u2013 5x \u2013 10x<\/span>2<\/span><\/p>\n

6 \u2013 12x<\/span>2<\/span><\/p>\n

Quadratic Function<\/b><\/h3>\n

A Quadratic Function can be represented in the form f(x) = ax<\/span>2<\/span> + bx + c, where a, b & c are numbers with \u201ca\u201d not equal to zero. This function is given so that you can plot it on a graph. The graph of a quadratic equation is a Parabola.<\/span><\/p>\n

Quadratic Equation Formulas & Basic Concepts<\/b><\/h3>\n

\u00a0<\/span>Now here are some basic concepts and important Quadratic Equation Formulas that are used to solve the questions. You should learn all the formulas by heart as without them you will not be able to solve the questions.<\/span><\/p>\n

The Quadratic Equation Questions can be solved using these two methods:<\/span><\/p>\n

    \n
  1. Standard Formula<\/li>\n
  2. Factorization Method<\/span><\/li>\n<\/ol>\n

    Let\u2019s understand these Quadratic Equation Concepts deeply.<\/span><\/p>\n

    \u00a0<\/span>Here we are explaining the traditional and shortcut methods to solve quadratic equation step by step<\/span><\/p>\n

    Check out this article on how to crack the banking exam with a good score here<\/strong>\u00a0<\/a><\/p>\n

    1. Standard Formula<\/b><\/h3>\n

    \u00a0<\/span>This formula is used to solve the Quadratic Equations easily and quickly. If an equation is in the form<\/span> ax\u00b2 + bx + c = 0<\/b>, then the solution will be<\/span> x = -b \u00b1 \u221a(b\u00b2 -4ac)\/2a.<\/b><\/p>\n

    You will get two values due to the \u201c+\u201d and \u201c-\u201d sign. To understand this method better, we have given an example of it.<\/span><\/p>\n

    \u00a0<\/span>Question:<\/b> Solve for x:- x<\/span>2<\/span> + 4x \u2013 8 = 0<\/span><\/p>\n

    Solution: <\/b>Here in this equation, using the standard formula, the values of a, b and c are 1, 4, and -8 respectively.<\/span><\/p>\n

    So, the solution will be:<\/span><\/p>\n

    x = -b \u00b1 \u221a(b\u00b2 -4ac)\/2a<\/span><\/p>\n

    x = -(-4) \u00b1 \u221a(4)<\/span>2<\/span> \u2013 4(1)(-8)\/ 2(1)<\/span><\/p>\n

    x = 4 \u00b1\u221a16 + 32\/2<\/span><\/p>\n

    = 2 \u00b1 \u221a48\/2<\/span><\/p>\n

    x = 2 \u00b1 2\u221a12\/2<\/span><\/p>\n

    x = 1 \u00b1 \u221a12<\/span><\/p>\n

    Therefore, the two values of the equation are (1 + \u221a12) and (1 \u2013 \u221a12).<\/span><\/p>\n

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    2. Factorization Method<\/strong><\/h3>\n

    In the following questions, two equations (a) and (b) are given. You have to solve both the equations and find the relation between x and y.<\/strong><\/p>\n

    Give answer (1) if x>y<\/b><\/p>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(2) if x>=y<\/b><\/p>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(3) if x<y<\/b><\/p>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(4) if x<=y<\/b><\/p>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(5) if x=y (or) the relation cannot be established.<\/b><\/p>\n

    Question 1\u00a0<\/b><\/p>\n

    1. 40x<\/b>2<\/b> – 47x + 12 =0<\/b><\/p>\n

    2. 5y2 – 51y + 54 =0<\/b><\/p>\n

    Normal method: (Factorization Method)<\/b><\/p>\n

    40x<\/b>2<\/b> – 47x + 12 =0<\/b><\/p>\n

    40x<\/span>2<\/span> – 32x-15x + 12 =0<\/span><\/p>\n

    8x(5x-4) – 3(5x-4) =0<\/span><\/p>\n

    (8x-3)(5x-4) =0<\/span><\/p>\n

    x= 3\/8 or 4\/5<\/span><\/p>\n

    5y2 – 51y + 54 =0<\/b><\/p>\n

    5y2 – 45y -6y + 54 =0<\/span><\/p>\n

    5y(y-9) -6(y-9) =0<\/span><\/p>\n

    (5y-6)(y-9) =0<\/span><\/p>\n

    y=6\/5 or 9<\/span><\/p>\n

    Clearly x<y and the hence correct option is (3)<\/b><\/p>\n

    Now the shortcut method<\/b><\/p>\n

    Step 1: Similar to factorization split the middle term such that its product gives the last term and sum give the middle term.\u00a0<\/span><\/p>\n

    \u00a0<\/span>For <\/span>\u00a040x<\/b>2<\/b> – 47x + 12 =0, <\/b>Sum = -47 and Product = 480. Hence the split terms are -32 and -15<\/span><\/p>\n

    Step 2: Now change the sign of split numbers, ie they become +32 and +15.<\/span><\/p>\n

    Step 3: Divide each of these numbers by the coefficient of x<\/span>2<\/span><\/p>\n

    In this case, it becomes 32\/40 and 15\/40<\/span><\/p>\n

    Therefore x= 4\/5, 3\/8 ie the same as the result obtained using the traditional method. Repeat the same process for finding the values of y and compare them with x. We will obtain the same result as above.\u00a0<\/span><\/p>\n

    \"tips<\/a><\/p>\n

    Question 2<\/b><\/p>\n

    1. <\/strong>12x<\/b>2<\/b> – 28x + 15 =0<\/b><\/p>\n

    2. 4y<\/b>2<\/b> – 20y + 21 =0<\/b><\/p>\n

    12x<\/b>2<\/b> – 28x + 15 =0<\/b><\/p>\n

    12x<\/span>2<\/span> -10x – 18x + 15 =0<\/span><\/p>\n

    2x(6x-5) -3(6x-5) =0<\/span><\/p>\n

    (2x-3)(6x-5)=0<\/span><\/p>\n

    x= 3\/2, 5\/6<\/span><\/p>\n

    4y<\/b>2<\/b> – 20y + 21 =0<\/b><\/p>\n

    4y<\/span>2 <\/span>-14y – 6y + 21 =0<\/span><\/p>\n

    2y(2y-7) – 3(2y-7) =0<\/span><\/p>\n

    ( 2y-3)( 2y-7) =0<\/span><\/p>\n

    y=3\/2, 7\/2<\/span><\/p>\n

    Here x<=y and the hence correct option is (4)<\/b><\/p>\n

    Now the shortcut method:\u00a0<\/b><\/p>\n

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    Let’s check out this method to find the solution based on the sign of coefficients in the equation<\/strong><\/h4>\n

    When we deal with the Quadratic Equations problem in Quantitative Aptitude, we should use time-saving tricks. The ideal & fast way to proceed, including the tricks, is given below:-<\/p>\n