Quadratic equations for IBPS PO Tricks & Shortcuts

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Quadratic equations for IBPS PO
How to crack Quadratic Equations for IBPS PO

Quadratic Equations for IBPS PO is one of the most important topics in the examination.

Quadratic equations is in general a very important topic for all banking examinations. This blog will give tips and tricks on cracking Quadratic equations for IBPS PO and SBI PO. Quant section of IBPS PO exam includes this type of questions. At least 5 out of 35 quant questions in IBPS PO prelims are on this topic. Quadratic equations that are asked in IBPS PO exams are easier to solve but sometimes they are calculation intensive. But, with regular practice, one can get adept at solving these types of questions.

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QUADRATIC EQUATIONS FOR IBPS PO:

Let’s start with the basic question structure of quadratic equations for IBPS PO exams:
Two quadratic equations of different variables will be given as shown below.

i) $x^2 + 4x + 3 = 0$

ii) $y^2+9y+20 =0$

Now the question asks us to solve for both x and y and to compare them.
Usually, the options will be as shown below.
a: X > Y
b: X < Y
c: X ≥ Y
d: X ≤ Y
e: X = Y or no relationship can be established between the two.

METHODS TO SOLVE QUADRATIC EQUATIONS:

We can solve these types of questions in two ways:

1) Directly solving for roots:

* By using the formula $x = \frac{-b \pm \sqrt(b^2-4ac)}{2a}$ we can solve for roots. But, one should be real quick while calculating the roots.

For example, let us take the equations mentioned above.

$x^2+4x+3 =0$ ——equation(1)

$y^2+9y+20=0$ —–equation(2)

Roots of the equation (1)

$x = \frac{-b \pm \sqrt(b^2-4ac)}{2a}$

==> $x= \frac{-4 \pm \sqrt(4^2-4*1*3)}{2*1}$
On solving we get $x= -3, -1.$

Roots of equation (2)
$y = \frac{-b \pm \sqrt(b^2-4ac)}{2a}$

==>$y = \frac{-9 \pm \sqrt(9^2-4*20*1)}{2a} $

On solving the above equation we get $y = -4, -5.$
On comparing the obtained roots of x and y we can clearly conclude that X is greater than Y.

*We can also calculate the roots of a quadratic equation by factorizing it.
For example, consider the below equation below

$x^2+2x-15=0$

==> $x^2 +5x -3x -15 =0$

==> $x(x+5)-3(x+5) =0$
==> $(x+5)(x-3) =0$
Therefore the roots of the equation are $x = -5, 3.$

2) Analyzing the signs of the equations:

By analyzing the signs of the given equations, we can infer the signs of their roots.
For example,
Let us consider the quadratic equation $ax^2 -bx+c =0$
By observing the signs, we can say that the sum of the roots of the equation (b/a) is positive. And the product of the roots of the equation (c/a) is also positive. Therefore both the roots must be positive.

And the equation of the form $ay^2+by+c =0$, the sum of the roots of the equation (-b/a) is negative. And the product of the roots of the equation (c/a) is positive, i.e., both the roots of the equation must be negative.
Since x is positive and y is negative irrespective of the values of a, b, c, x is always greater than Y.

Solving quadratic equations through this method saves a lot of time. But, the drawback of this method is, we can only analyze equations which are among the forms mentioned above.

LINEAR MIXED VARIABLE EQUATIONS:

Sometimes instead of giving quadratic equations, they give two two-variable linear equations and ask us to solve for those two variables and compare them. Solving linear equations is easier compared to solving quadratic equations

An easier way to solve these linear equations is by simplifying the given equations such that the coefficients of a particular variable in both the equations become equal. This can be done by multiplying equation/s with constants.

For example, let’s take a look at the following two linear equations:
3x – y =1 —- equation (1)
4x -2y +4 = 0 —– equation (2)
Let’s solve the above two equations by equating the coefficients of y in both the equations.
Equation (1) * (-2) ==> -6x + 2y = -2
Equation (2) ==> 4x -2y +4 =0
Adding both these equations we get x =3.
On putting x=3 either in equation (1)/(2) we get y = 8.
Therefore X is less than Y.

Now that we have discussed the basics on how to solve quadratic equations for IBPS PO, let’s take a look at the following solved examples.

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EXAMPLES:

Instructions: In each of these questions, two equations (I) and (II) are given which contain information about two variables X and Y. Solve both the equations and select the appropriate answer based on the relationship between X and Y.

Example 1:

$14x^2-55x+21=0$

$6y^2-83y+286=0$

a: X > Y
b: X < Y
c: X ≥ Y
d: X ≤ Y
e: X = Y or no relationship can be established between the two.

Explanation:
Equation (1)

$14x^2 -55x+21=0$

==> $14x^2 -6x-49x+21=0$
==> $2x(7x-3)-7(7x-3) =0$
==> $(7x-3)(2x-7)=0$
Therefore roots of equation 1 are 7/2, 3/7.
Similarly, equation (2)

$6y^2-83y+286=0$

==> $6y^2-44y-39y+286=0$
==> $2y(3y-22) -13(3y-22) =0$
==> $(2y-13)(3y-22)=0$
Therefore roots of the equation 2 are 13/2, 22/3.
It is clear that x < Y.
Therefore, the correct option to choose is B.

Example 2:

$3x^2+53x+111=0$

$y^2-8y+2=0$

 

a: X > Y
b: X < Y
c: X ≥ Y
d: X ≤ Y
e: X = Y or no relationship can be established between the two.
Explanation:
Equation (1):
Observe the coefficients in the given equation,
Sum of the roots = -53/3
Product of the roots = 111/3
Therefore both the roots must be negative.
Equation (2):
Sum of the roots = 8
Product of the roots = 2
Therefore both the roots must be positive.
So, X < Y
Therefore the correct option to choose is B.
So, this is all about quadratic equations for IBPS PO; I would recommend practicing regularly in order to avoid silly mistakes.

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Do check an excellent blog on Number Systems for SBI PO exam here.

Also check out a very helpful blog on High Level Reasoning for SBI PO here.

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