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Quadratic Equation Tricks & Tips

Quadratic Equation Tricks & Tips


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What is Quadratic Equation?

Any equation having in the form of ax2+bx+c=0, is a quadratic equation in where x represents an unknown. a, b & c represents the real numbers such that a is not equal to 0.

In this Equation:

a is the coefficient of x2

b is the coefficient of x

c is the constant term

How to solve Single Variable Quadratic Equation ?

In single variable quadratic equation we will be given one quadratic equation in the form of  ax2+bx+c=0 & we have to find the value of that unknown variable x.

In quadratic Equations We have the following options to choose from :-

  1. X >Y
  2. Y >X

iii. X >= Y

  1. Y>= X
  2. X = Y or relationship cannot be established.

 

(i) Now, Suppose on solving the quadratic equation we get X = -1, 2 And Y = -2, -3

On putting these values on number line we will see that X lies on the right of Y. Hence, We can say that X>Y or Y<X.

(ii) Now, Suppose if after solving the equation we get X = -1 , 2 and Y= 3,4.

On putting these values on number line we see that Y lies to the right of X without touching X at any point. Hence, We can say that Y>X or X<Y.

(iii) If X= 3,4 and Y= 2,3. Putting this on number line we get X to the right of Y but touching Y at one single point i.e., 3. Hence, X>=Y or Y<=X.

(iv) If X=3,4 and Y=4,5. Putting these values on number line we will see that Y lies to the right of X but touching X at 4. Hence, Y>=X or X<=Y.

(v) Now, Suppose X= 2,4 and Y= 3,5. Putting these values on number line we will see that X touches Y at all the points between 3 and 4 i.e., at 3, 3.1, 3.2, 3.3 etc. upto 4 hence, in this case we cannot relate X and Y and Hence, Will answer as Relationship cannot be established.

Also suppose if y=7 and X= 4,8 then X<y also X>y . Hence, in this case also no relationship can be established.

Example:

x2+13x+40=0 —-(1)

y2+7y+12=0 —– (2)

First Solve the Equation (1)

Therefore, +13 = (+5) + (+8)

Also, +40 = (+5) * (+8)

Now Change the signs & Divide them with the coefficient of x2 i.e with +1

Therefore, we get

-5/1= -5 & -8/1= -8

Thus, -5 & -8 are the required values of x.

Now similarly Solve the 2nd Equation,

+7 = (+3) + (+4)

Also +12 = (+3) * (+4)

Now Change the signs & Divide them with the coefficient of y2 i.e with +1

-3/1=-3 & -4/1=-4

Thus, -3 & -4 are the required values of y.

Therefore Now we have x= -5, -8

y= -3, -4

Therefore, by comparing both the values we can say that x < y

Short Tricks to Solve the Quadratic Equation :

+      + –          
–       + +        +
+       – –         +
–        – +        –

Remember this diagram friends to solve the Questions of quadratic equation shortly in the exam but before exam you will need to practice more and more question to get speed and also to get familiar with the tricks.

  1. 1. I.5×2– 87x + 378 = 0                 3y2– 49y + 200 = 0

I.5×2 – 45x – 42x + 378 = 0

or,5x(x – 9) – 42(x – 9) = 0

or,(5x – 42) (x – 9) = 0

x = 9,5

II.3y2 – 24y – 25y + 200 = 0

or,3y(y – 8) –25(y – 8) = 0

or,(y – 8) (3y – 25) = 0

y = 8,3

So , Answers : x > y

Now for this question we can solve this in less than thirty second how lets see,

Step-1. First take the Equation-I means equation with  variable x.

It is in the form of   ax2  +   (-)b x + c.

Step-2. See the Sign of b & c means sign of coefficient of x variable and sign of last one,

It is first negative(-) and second positive(+)

Step -3.From above table we can say that both the roots of x will be positive.

Step-4.Make Calculation 378*5=1890 first divide by 2 then 3 and then by 5 and follow above rule, so for this it will be 45 and 42 and now you don’t need to write another two to three lines of taking common and then making in the forms of roots. This is your answer because we know from table that both root have positive sign.

Step-5 For those question in which coefficient of X2 is not one like in this question you will need to divide both the roots by (a).

So answer will be :-9,5

 

Example some quadratic equation:
Example 1:

(i) 6X^2 +11X + 3 = 0

(ii) 6X^2 + 10X +4= 0

Answer :
(i) Shortcut tricks :
This equation +6 is coefficient of x2 .
+ 11 is coefficient of x
+3 is constant term.
Step 1: we multiply (+6) x (+3) = +18 = 2x3x3
Step 2: we break 2x3x3 in two parts such that addition between them is 11.
So, 2x3x3 = 2 x 9. Also,  9 +2 = 11

So , +9 and +2 = Sum of is +11 .

Step 3: Change the sign of both the factors , So +9 = -9 and +2 = -2 .
and divide by coefficient of x2 , So we get -9 / 6 = -3 / 2 and -2 / 6 = – 1 / 3 .

Therefore, X = -3/2 , -1/3

Similarly Solving (ii),

6×4= 24 = 2x2x2x3

Break 2x2x2x3 into two parts such that their sum becomes 10.

2×2 and 2×3 are two parts of 24 whose sum is 10.

Now, Change the sign of both the factors  and divide by coefficient of X^2.

So, +4 = > -4/6

And +6 = > -6/6

Y = –2/3 , -1

So, in all X= -1.5, -0.33

And, Y = -0.667, -1

Now, Putting these values on Number line we get to knw that from -0.667 to -1

The values of X and Y coincides .

Hence, as X and Y coincides at more than 1 point,

We can Say X=Y or Relationship between X and Y cannot be established.

 

QUESTION

(i)           X2 – 11X + 28 = 0

(ii)         Y2 – 15Y + 56 = 0

GIVEN

In equation (i)

Sum of Root (SR) = 11

Product of Root (PR) = 28

Similarly in eq. (ii)

SR= 15

PR= 56

SOLUTION:

NORMAL METHOD:

(i). X2– 11X + 28 = 0

Now SR = -11 can be written as (-7-4 = -11)

So X2 – 7X – 4X + 28 = 0

Consider the first 2 terms and take the common term outside i.e., X here

X(X – 7) – 4X + 28 = 0

Similarly consider the last 3 terms and take the common term outside i.e., -4 here

X(X – 7) – 4(X – 7) = 0

(X – 7) (X – 4) = 0

ThereforeX = 7, 4

(ii). Y2– 15Y + 56 = 0

Now SR = -15 can be written as (-7-8 = -15)

So Y2 – 7Y – 8Y + 56 = 0

Consider the first 2 terms and take the common term outside i.e., Y here

Y(Y – 7) – 8Y + 56 = 0

Similarly consider the last 3 terms and take the common term outside i.e., -8 here

Y(Y – 7) – 8(Y – 7) = 0

(Y – 7) (Y – 8) = 0

ThereforeY = 7, 8

We have calculated the values of X and Y, now we have to compare the values with each other to deduce the relation between them

TakeX = 7, compare it with both the values of Y = 7, 8

We get, X = 7 is equal to Y = 7 i.e.,X=Y

X = 7 is smaller than Y = 8 i.e.,X<Y

Similarly TakeX = 4, compare it with both the values of Y = 7, 8

We get, X = 4 is smaller than Y = 7 i.e.,X<Y

X = 4 is smaller than Y = 8 i.e.,X<Y

So the relation between X and Y is given by both X = Y and X<Y i.e.,X≤Y

ThereforeAnswer is (4) if X≤Y

ALTERNATE METHOD:

If the givenSR is–ve then consider it as+ve

If the givenSR is+ve then consider it as–ve

Split the PR into its divisible numbers such that when the numbers are added or subtracted we get the SR

Here 7 × 4 = 28 (PR)

And 7 + 4 = 11 (SR)

Here 7 × 8 = 56 (PR)

And 7 + 8 = 15 (SR)

Therefore from both the equationsX = 7, 4 andY = 7, 8

TakeX = 7, compare it with both the values of Y = 7, 8

We get, X = 7 is equal to Y = 7 i.e.,X=Y

X = 7 is smaller than Y = 8 i.e.,X<Y

Similarly TakeX = 4, compare it with both the values of Y = 7, 8

We get, X = 4 is smaller than Y = 7 i.e.,X<Y

X = 4 is smaller than Y = 8 i.e.,X<Y

So the relation between X and Y is given by both X = Y and X<Y i.e.,X≤Y

Example 1:P2+13P+40=0

Q2+7Q+12=0

Step1: let us take equ 1. P2+13P+40=0 in this equation coefficient of p, 13 should be split into two numbers in such a way that multiplication of both numbers should be equal to constant term 40 and addition of numbers should be equal to 13

13 can be split into (1,12) (2,11) (3,10) (4,9) (5,8) (6,7)

In these combination 5 and 8 only can give 40 while multiplying, so this is the number we are searching for,and since the coefficient of P2 is 1 and there is no negative sign in the equation, we can directly write value of P by simply changing the sign

P= -5,-8

( just for reference Actual procedure is  P2+5P+8P+40=0

P(P+5)+8(P+5)=0

(P+5)(P+8)=0

P=-5,-8 )

Step 2: Now equ2. Q2+7Q+12=0 similar process applicable for this equation to find Q, here coefficient of Q should be split into two numbers and multiplication of the numbers should give 12

7 can be split up into (1,6) (2,5) (3,4)

Combination of 3 and 4 alone satisfy our need i.e. giving 7 and 12 while adding and multiplying the numbers respectively, since there is no negative sign in the equation,we can directly write value of Q by changing sign .

Q=-3, -4

P= -5,-8

Obviously P < Q

Example 2: 2p2+12p+16=0

2q2+14q+24=0

Step 1: let us take equ1 since coefficient of p2 is 2,we have to multiply 2 with constant number 16 , now as usual 12 should be split up into two numbers and multiplication of the numbers should give ( 2*16=) 32 .

eg2

4+8 =12

4*8=32

12 can be split up into 4 and 8,

now change the sign of numbers and divide it by 2 since the coefficient of p2 is 2 ,

Thus the value of  p = -2,-4

Step 2: now take equ2. And follow the same procedure multiply 2 with constant number 24 . So addition of two numbers should be 14 and multiplication of numbers should be 48 (24*2)

eg2_2

6+8 =14  ;  6*8=48

Numbers are 6 and 8, now divide by 2, since the coefficient of qis 2 so the value of q is -3,-4

here, one of the numbers of p and q are same but other number of p is greater than other number of q

so ans is p ≥ q

If you have doubt in finding which one is greater, use this technique.

Number which is 1st from right hand side is greater one.

ruler

Example 3 x2-x-6=0

2y2+13y+21

Step1: Here constant number and coefficient of x are negative so the combination of numbers will be positive and negative.

eg3_1

Combination is 2,-3

Value of x=-2 , 3

Step2: proceed with equation 2

eg3_2

Combination is (6,7) ,

since the coefficient of y2 is 2 divide the value of y by 2  

Y=  -(6/2) , -(7/2)

Y= -3 , -3.5

Relation is X > Y

Example 412x2+11x+12=10x2+22x

13y2-18y+3=9y2-10y

Step1: convert this into normal quadratic equation form

2x2-11x+12=0

4y2-8y+3=0

Step2: now as usual normal procedure , here coefficient of x is negative and constant term is positive so both numbers will be negative

eg4_1

Combination is -8,-3

And value of x= 4, 3/2

Step3: proceed with equ 2

eg4_2

Combination is (-2,-6)

since coefficient of yis 4 divide the value of y by 4 ,

And value of y = ½ , 3/2

Relation is x ≥ y

Example 5: 18/x2 + 6/x -12/x2 = 8/x2

Y3+9.68+5.64 =16.95

Multiply xin equ1 it becomes 18+6X -12 =8

Solving this equations we get x = 1/3 = 0.33

Solving equ2 we get y= 1.63

Y = 1.17

Relation is x < y

Example 6: (x+18)1/2 = (144)1/2 – (49)1/2

Y+409 = 473

By solving equ1, we get x = 7

Solving equ2 we get y = ± 8

Relation cannot be formed between x and y since x is both highest and lowest one.

TIPS to find combination:

  1. If the coefficient of x or y and constant term is negative, then one number will be positive and other will be negative.
  2. If the coefficient of x or y and constant term is positive, combination will be positive.
  3. If the coefficient of x or y and constant term are positive and negative then the combination of numbers will be both positive and negative.

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

1. Write down the table (given below) before exam starts, in your rough sheet, for use during exam, Analyse the (+, -) signs in the problem, and refer to the table of signs,

2. Write down the new (solution) signs, and see if a solution is obtained instantly. If not, then go to step 3.

3. Obtain the two possible values for X & Y, from both the equations,

4. Rank the values and get the solution,

STEP 1

Firstly, when you enter the exam hall, you need to write down the following master table in your rough sheet instantly (only the signs):-

How to Solve Quadratic Equation? – Tips & Tricks

Quadratic

NOTE: The above table should be entered before the exam starts, in rough sheet. Because, this will save time, and you can refer to it instantly later on.

STEP 2

Try solving the question instantly if possible, by checking whether + or – values can bring us to a conclusion.

See an example No. 1:- 

x2 + 7x + 12 = 0

y2 – 5y + 6 = 0

The signs of X’s equation are + and +, which means their solution is – and. Both negative values. (Refer to the table)

The signs of Y’s equation are – and +, which means their solution values are + and +. Both positive values.

Obviously, X’s possible values are both negative… And Y’s possible values are both positive.

Obviously, solution is X < Y. We found out just by looking at the signs which will take hardly 5 seconds.

So, any question with signs as “+, +” and “-, +” can be instantly solved, (unless its an unsolvable exception like x2 – 6x + 16 = 0, which can have no real integer as answer. But this is a very rare occasion).

See another example, Example No. 2:-

x2 + 21x – 32 = 0

y2 + 7y – 12 = 0

The question’s signs are + and -, for both X & Y equations. Which means X and Y’s values can be both positive or negative, as per master table…. Answer is an instant CND (Can Not Determine). This takes hardly 5 seconds.

Similarly, any question’s 2 equations with all 4 symbols as negative (-, -), will have each variable’s value either a positive or negative, so its again a CND.

STEP 3

If you visit Step 3, it means the question is not an instantly solved in Step 2. Here, we write the 2 possible signs of each variable anyways, as plus or minus, in our rough sheet.

We must find the possible values of the variables. Remember, from the 2 symbols (+ or -) derived from the master table, the first symbol connects to the bigger value, and second symbol connects to the smaller value, numerically.

Example 3:-

x2 – 12x + 32 = 0

y2 – 7y + 12 = 0

The question’s symbols are -, + which means both values will be positive for each variable, x and y.

x2 – 4x – 8x + 32 = 0

X (x – 4) – 4 (x – 8) = 0

X = 4, X = 8

y2 – 7y + 12 = 0

y2 – 4y – 3y + 12 = 0

y (y – 4) – 3 (y – 4) = 0

Y= 4, Y = 4

So, X >= Y

This is a very basic example. Let’s move on to fractional example.

Example 4 is given below:-

2x2 + 3x – 9 = 0

2y2 – 11y + 15 = 0

+, – means X will be (-) or (+).

-, + means Y will be (+) or (+)

We have to solve it as we can’t decide instantly.

Now, due to a variation of a factor attached to squared values, the change in method is that we’ll have to multiply variable less value (9) with factor of the squared variable (2), to get 18 for X’s equation.

We factorize 18 as 2 x 3 x 3 and reduce it to 2 values (which add/subtract to 3) which is 6 x 3…. Now, divide both values by 2, because that is factor of x2.

So, x = -6/2 or +3/2… In other words, x = -3 or 1.5.

Similarly for Y, we know the signs are + and -, which means both values are positive.

And 15 x 2 = 30. (Again, this is done because y2 has a multiplying factor of 2)

30 = 2 x 3 x 5. (we’ll have to multiply 2 values of these, to arrive at final 2 values which add upto 11, which are 6 and 5)

11 = 6 + 5. Now, Divide both 6 and 5 by 2 each, to get respective possible values of y.

So, x = -6/2, 5/2

y = 3 or 2.5.

STEP 4

Rank all the 4 values carefully, considering – as lower values and + as higher, as they normally are in mathematics.

For example,

“-5” will get rank 4,

“-0.8” will get rank 3,

“0.4” will get rank 2,

and “4” will get rank 1.

Write the ranks in your rough sheet, besides the values. Check the ranks and see what conclusion can be derived.

If there’s no tie up/ draw among the ranks, then the answer is easy to find…

Quadratic 2

So, as you can see from the table above, when all 4 values are different, and if the ranks 1 and 2 are together, they make it a senior variable, otherwise, we can’t determine and it becomes a CND.

When they are tied up, then we’ll have to work it out on rank to rank comparison basis to find a final solution, like below. But this we do only in mind:-

Let’s carry Example 3 from the Step 3 mentioned before…

X = -8, -4

Y = -4, -3

X’s ranks would be 4,2

Y’s ranks would be 2,1. (assuming that ranks 2 & 3 are combined to get rank 2)

We compare them by “Best vs Worst” method this way:-

Quadratic 3

If we ever have to combine a X>Y with X<Y, then its obviously CND. Rest you guys can figure out. Remember to do this Step 4 in mind itself, writing only the ranks on rough sheet and nothing else. Don’t make such tables there when the clock is running out…

Let’s solve Example 4 further from where we left:-

x = -6/2, 5/2

y = 3 or 2.5.

X’s ranks are= 4, 2

Y’s ranks are= 1, 2.

We compare best to worst, and solve it in mind itself to get Y >= X.

Now let us check some special cases of inequality.

SPECIAL CASES:-

1. SQUARES – Let us take few examples of non equation values. See example No. 5:-

x2 = 1600

y2 = 1600

Answer is CND. Because, a square would always mean possibility of a positive as well as negative value,

x = 40, -40

y = 40, -40

Hence, any 2 squares or their variations are always CND, irrespective of the values.

 NOTE: Root of a positive value is always positive. If you have to root a negative value, then the question is wrong, and hence, answer is a CND.

See example No. 6:-

x2 – 243 = 468

y2 + 513 = 1023

Don’t touch the pen. Its a CND. Because, ultimately we’ll get a perfect/ imperfect square, which is always a CND.

2. LINEAR MIXED VARIABLE EQUATIONS

Example 7:-

5x + 4y = 82

4x + 2y = 64

One way to proceed is to equate either the “X”or “Y” part of both equations by multiplication any one equation with a positive integar.

Here, best way would be to double the second equation to form 8x + 4y = 128. We can now substract second equation from first,

8x + 4y  = 128

–  5x + 4y  = 82

————————

3x   = 36

removing 4y from both sides easily to get X=12,

4y = 82 – 5x = 82 – 60 = 22,

y = 5.5,

There’s mostly no positive negative confusion in these cases. So, X > Y is right here.

Practice Problems On Quadratic Equation


  1. x² – 34x + 288 = 0
    y² – 28y + 192 = 0
    A. X > Y
    B. X < Y
    C. X ≥ Y
    D. X ≤ Y
    E. X = Y or relation cannot be established

    Answer & Explanation
    C. X ≥ Y
    Explanation:

    x² – 34x + 288 = 0
    x = 18, 16
    y² – 28y + 192 = 0
    y = 12, 16
  2. x² – 26x + 168 = 0
    y² – 32y + 252 = 0
    A. X > Y
    B. X < Y
    C. X ≥ Y
    D. X ≤ Y
    E. X = Y or relation cannot be established

    Answer & Explanation
    D. X ≤ Y
    Explanation:

    x² – 26x + 168 = 0
    x = 12, 14
    y² – 32y + 252 = 0
    y = 14, 18
  3. x² + 26x + 168 = 0
    y² + 23y + 132 = 0
    A. X > Y
    B. X < Y
    C. X ≥ Y
    D. X ≤ Y
    E. X = Y or relation cannot be established

    Answer & Explanation
     D. X ≤ Y
    Explanation:

    x² + 26x + 168 = 0
    x = -12, -14
    y² + 23y + 132 = 0
    y = -12, -11
  4. x² – 28x + 195 = 0
    y² – 30y + 216 = 0
    A. X > Y
    B. X < Y
    C. X ≥ Y
    D. X ≤ Y
    E. X = Y or relation cannot be established

    Answer & Explanation
     E. X = Y or relation cannot be established
    Explanation:

    x² – 28x + 195 = 0
    x = 15, 13
    y² – 30y + 216 = 0
    y = 18, 12
  5. (x – 19)² = 0
    y² = 361
    A. X > Y
    B. X < Y
    C. X ≥ Y
    D. X ≤ Y
    E. X = Y or relation cannot be established

    Answer & Explanation
     C. X ≥ Y
    Explanation:

    x² – 38x + 361 = 0
    x = 19, 19
    y² = 361
    y = ±19
  6. x² – 44x + 448 = 0
    y² – 28y + 195 = 0
    A. X > Y
    B. X < Y
    C. X ≥ Y
    D. X ≤ Y
    E. X = Y or relation cannot be established

    Answer & Explanation
     A. X > Y
    Explanation:

    x² – 44x + 448 = 0
    x = 28, 16
    y² – 28y + 195 = 0
    y = 13, 15
  7. x² – 26x + 168 = 0
    y² – 34y + 285 = 0
    A. X > Y
    B. X < Y
    C. X ≥ Y
    D. X ≤ Y
    E. X = Y or relation cannot be established

    Answer & Explanation
     B. X < Y
    Explanation:

    x² – 26x + 168 = 0
    x = 12, 14
    y² – 34y + 285 = 0
    y = 15, 19
  8. x² – 38x + 352 = 0
    y² – 25y + 154 = 0
    A. X > Y
    B. X < Y
    C. X ≥ Y
    D. X ≤ Y
    E. X = Y or relation cannot be established

    Answer & Explanation
     A. X > Y
    Explanation:

    x² – 38x + 352 = 0
    x = 22, 16
    y² – 25y + 154 = 0
    y = 11, 14
  9. x² = 121
    y² – 46y + 529 = 0
    A. X > Y
    B. X < Y
    C. X ≥ Y
    D. X ≤ Y
    E. X = Y or relation cannot be established

    Answer & Explanation
     B. X < Y
    Explanation:

    x² = 121
    x = 11, – 11
    y² – 46y + 529 = 0
    y = 23, 23
  10. x² – 31x + 234 = 0
    y² – 34y + 285 = 0
    A. X > Y
    B. X < Y
    C. X ≥ Y
    D. X ≤ Y
    E. X = Y or relation cannot be established

    Answer & Explanation
    E. X = Y or relation cannot be established
    Explanation:

    x² – 31x + 234 = 0
    x = 13, 18
    y² – 34y + 285 = 0
    y = 15, 19

 

  • x² – 32x + 252 = 0
    y² – 28y + 192 = 0
    A. X > Y
    B. X < Y
    C. X ≥ Y
    D. X ≤ Y
    E. X = Y or relation cannot be established

    Answer & Explanation
     E. X = Y or relation cannot be established
    Explanation:

    x² – 32x + 252 = 0
    x = 18, 14
    y² – 28y + 192 = 0
    y = 12, 16
  • x² – 32x + 247 = 0
    y² – 22y + 117 = 0
    A. X > Y
    B. X < Y
    C. X ≥ Y
    D. X ≤ Y
    E. X = Y or relation cannot be established

    Answer & Explanation
     C. X ≥ Y
    Explanation:

    x² – 32x + 247 = 0
    x = 13, 19
    y² – 22y + 117 = 0
    y = 13, 9
  • x² + 29x + 208 = 0
    y² + 19y + 78 = 0
    A. X > Y
    B. X < Y
    C. X ≥ Y
    D. X ≤ Y
    E. X = Y or relation cannot be established

    Answer & Explanation
     D. X ≤ Y
    Explanation:

    x² + 29x + 208 = 0
    x = -13, -16
    y² + 19y + 78 = 0
    y = -13, -6
  • x² – 22x + 105 = 0
    y² – 27y + 162 = 0
    A. X > Y
    B. X < Y
    C. X ≥ Y
    D. X ≤ Y
    E. X = Y or relation cannot be established

    Answer & Explanation
     E. X = Y or relation cannot be established
    Explanation:

    x² – 22x + 105 = 0
    x = 15, 7
    y² – 27y + 162 = 0
    y = 18, 9
  • (x – 18)² = 0
    y² = 324
    A. X > Y
    B. X < Y
    C. X ≥ Y
    D. X ≤ Y
    E. X = Y or relation cannot be established

    Answer & Explanation
     C. X ≥ Y
    Explanation:

    x² – 36x + 324 = 0
    x = 18, 18
    y² = 324
    y = ±18
  • x² – 41x + 400 = 0
    y² – 29y + 210 = 0
    A. X > Y
    B. X < Y
    C. X ≥ Y
    D. X ≤ Y
    E. X = Y or relation cannot be established

    Answer & Explanation
     A. X > Y
    Explanation:

    x² – 41x + 400 = 0
    x = 25, 16
    y² – 29y + 210 = 0
    y = 14, 15
  • x² – 25x + 156 = 0
    y² – 32y + 255 = 0
    A. X > Y
    B. X < Y
    C. X ≥ Y
    D. X ≤ Y
    E. X = Y or relation cannot be established

    Answer & Explanation
     B. X < Y
    Explanation:

    x² – 25x + 156 = 0
    x = 12, 13
    y² – 32y + 255 = 0
    y = 15, 17
  • x² – 35x + 294 = 0
    y² – 23y + 132 = 0
    A. X > Y
    B. X < Y
    C. X ≥ Y
    D. X ≤ Y
    E. X = Y or relation cannot be established

    Answer & Explanation
     A. X > Y
    Explanation:

    x² – 35x + 294 = 0
    x = 21, 14
    y² – 23y + 132 = 0
    y = 11, 12
  • x² = 64
    y² – 34y + 289 = 0
    A. X > Y
    B. X < Y
    C. X ≥ Y
    D. X ≤ Y
    E. X = Y or relation cannot be established

    Answer & Explanation
     B. X < Y
    Explanation:

    x² = 64
    x = 8, – 8
    y² – 34y + 289 = 0
    y = 17, 17
  • x² – 29x + 208 = 0
    y² – 32y + 255 = 0
    A. X > Y
    B. X < Y
    C. X ≥ Y
    D. X ≤ Y
    E. X = Y or relation cannot be established

    Answer & Explanation
    E. X = Y or relation cannot be established
    Explanation:

    x² – 29x + 208 = 0
    x = 13, 16
    y² – 32y + 255 = 0
    y = 15, 17

Directions: In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly-

  1. I. 4x2 – 29x + 45 = 0,
    II. 4y2 – 17y + 18 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established

    View Answer
     Option C
    Solution: 

    4x2 – 29x + 45 = 0
    4x2 – 20x – 9x + 45 = 0
    Gives x = 9/4, 5
    4y2 – 17y + 18 = 0
    4y2 – 8y – 9y + 18 = 0
    Gives y = 2, 9/4
  2. I. 3x2 – 13x – 30 = 0,
    II. 2y2 – 25y + 78 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established

    View Answer
     Option D
    Solution: 

    3x2 – 13x – 30 = 0
    3x2 – 18x + 5x – 30 = 0
    Gives x = -5/3, 6
    2y2 – 25y + 78 = 0
    2y2 – 12y – 13y + 78 = 0
    Gives y = 6, 13/2
  3. I. 3x2 – 20x + 32 = 0,
    II. 3y2 – 29y + 56 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established

    View Answer
     Option E
    Solution: 

    3x2 – 20x + 32 = 0
    3x2 – 12x – 8x + 32 = 0
    Gives x = 8/3, 4
    3y2 – 29y + 56 = 0
    3y2 – 21y – 8y + 56 = 0
    Gives y = 8/3, 7
  4. I. 3x2 – 16x – 35 = 0,
    II. 3y2 – 23y + 40 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established

    View Answer
     Option E
    Solution: 

    3x2 – 16x – 35 = 0
    3x2 – 21x + 5x – 35 = 0
    Gives x = -5/3, 7
    3y2 – 23y + 40 = 0
    3y2 – 15y – 8y+ 40 = 0
    Gives y= 8/3, 5
  5. I. 2x2 – 23x + 65 = 0,
    II. 3y2 + 2y – 16 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established

    View Answer
     Option A
    Solution: 

    2x2 – 23x + 65 = 0
    2x2 – 10x – 13x + 65 = 0
    Gives x = 5, 13/2
    3y2 + 2y – 16 = 0
    3y2 – 6y + 8y – 16 = 0
    Gives y= -8/3, 2
  6. I. x2 – 78 = 91,
    II. √3 y = √432
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established

    View Answer
     Option E
    Solution: 

    x2 – 78 = 91
    x2 = 169
    Gives x = -13, 13
    √3 y = √432
    y = √432/√3 = √144
    Gives y = 12
    Now when x = -13, y > x
    and when x = 13, y < x
    So relation cannot be established
  7. I. 3x2 + 17x + 10 = 0,
    II. 3y2 + 14y – 5 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established

    View Answer
     Option E
    Solution: 

    3x2 + 17x + 10 = 0
    3x2 + 15x + 2x + 10 = 0
    Gives x = -5, -2/3
    3y2 + 14y – 5 = 0
    3y2 + 15y – y – 5 = 0
    Gives y = -5, 1/3
  8. I. 2x2 – 13x + 15 = 0,
    II. 2y2 + 5y – 12 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established

    View Answer
     Option C
    Solution: 

    2x2 – 13x + 15 = 0
    2x2 – 10x – 3x + 15 = 0
    Gives x = 3/2, 5
    2y2 + 5y – 12 = 0
    2y2 + 8y -3y – 12 = 0
    Gives y= -4, 3/2
  9. I. 2x2 – 3x – 35 = 0,
    II. 3y2 + 11y + 6 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established

    View Answer
     Option E
    Solution: 

    2x2 – 3x – 35 = 0
    2x2 – 10x + 7x – 35 = 0
    Gives x = -7/2, 5
    3y2 + 11y + 6 = 0
    3y2 + 9y +2y + 6 = 0
    Gives y= -3, -2/3
  10. I. 3x2 + 19x + 20 = 0,
    II. 3y2 – 7y – 6 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established

    View Answer
    Option B
    Solution: 

    3x2 + 19x + 20 = 0
    3x2 + 15x + 4x + 20 = 0
    Gives x = -5, -4/3
    3y2 – 7y – 6 = 0
    3y2 – 9y + 2y – 6 = 0
    Gives y= -2/3, 3

Directions: In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly-

  1. I. 2x2+25x + 78 = 0,
    II. 3y2 + 23y + 30 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established

    View Answer
    Option D
    Solution: 

    2x2+ 25x + 78 = 0,
    2x2+12x + 13x + 78 = 0
    Gives x = -13/2, -6
    3y2 + 23y + 30 = 0
    3y2 + 18y + 5y + 30 = 0
    Gives y = -6, -5/3
    Put all values on number line and analyze the relationship
    -13/2…. -6….-5/3
  2. I. 2x2 + 7x – 15 = 0,
    II. 3y2 + 11y ¬– 20 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established

    View Answer
    Option E
    Solution: 

    2x2 + 7x – 15 = 0
    2x2 + 10x – 3x – 15 = 0
    Gives x = -5, 3/2
    3y2 + 11y ¬– 20 = 0
    3y2 + 15y ¬– 4y – 20 = 0
    Gives y = -5, 4/3
    Put all values on number line and analyze the relationship
    -5.…. 3/2….4/3
    When x = 3/2, it is both > y(-5) and < y(4/3)
  3. I.3x2– 11x + 6 = 0,
    II. 3y2 + 11y ¬– 20 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established

    View Answer
    Option E
    Solution: 

    3x2– 11x + 6 = 0
    3x2–9x – 2x + 6 = 0
    Gives x = 2/3, 3
    3y2 + 11y ¬– 20 = 0
    3y2 + 15y ¬– 4y – 20 = 0
    Gives y = -5, 4/3
  4. I.3x2 + 17x + 20 = 0,
    II. 3y2– 4y – 15 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established

    View Answer
    Option D
    Solution: 

    3x2 + 17x + 20 = 0
    3x2 + 12x + 5x + 20 = 0
    Gives x = -4,-5/3
    3y2– 4y – 15 = 0
    3y2– 9y + 5y – 15 = 0
    Gives y= -5/3, 3
  5. I. 5x2– 19x + 12 = 0,
    II. 5y2 +6y – 8 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established

    View Answer
    Option C
    Solution: 

    Explanation:
    5x2 – 19x + 12 = 0
    5x2 – 19x + 12 = 0
    Gives x = 4/5, 3
    5y2 + 6y – 8 = 0
    5y2 + 10y – 4y – 8 = 0
    Gives y= -2, 4/5
  6. I.3x2– 10x – 8 = 0,
    II. 2y2 + 13y + 21 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established

    View Answer
    Option A
    Solution: 

    3x2– 10x – 8 = 0
    3x2– 12x + 2x – 8 = 0
    Gives x = -2/3, 4
    2y2 + 13y + 21 = 0
    2y2 + 6y + 7y + 21 = 0
    Gives y = -7/2, -3
  7. I. 4x2–15x + 9 = 0,
    II. 2y2 – 15y + 27 = 0
    A) x > y
    B) x< y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relationship cannot be determined

    View Answer
    Option D
    Solution: 

    4x2 –15x + 9 = 0
    4x2 –12x – 3x + 9 = 0
    Gives x = 3/4, 3
    2y2 – 15y + 27 = 0
    2y2 – 6y – 9y + 27 = 0
    So y = 3, 9/2
  8. I. 3x2 –14x + 8 = 0,
    II. 2y2 – 3y ¬– 20 = 0
    A) x > y
    B) x< y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relationship cannot be determined

    View Answer
    Option E
    Solution: 

    3x2 –14x + 8 = 0
    3x2 –12x – 2x + 8 = 0
    Gives x = 2/3, 4
    2y2 – 3y ¬– 20 = 0
    2y2 – 8y + 5y ¬– 20 = 0
    So y = -5/2, 4
    When x = 2/3, x > y(-5/2) and also x < y (4), so relationship cannot be determined
  9. I. 4x2– (1– 8√2)x– 2√2 = 0
    II. 5y2 + (1 + 5√2)y + √2 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established

    View Answer
    Option E
    Solution: 

    4x2– (1 – 8√2)x – 2√2 = 0
    (4x2 – x) + (8√2x – 2√2) = 0
    x (4x – 1) + 2√2 (4x – 1) = 0
    So x = 1/4, -2√2 (-2.82)
    5y2 + (1 + 5√2)y + √2 = 0
    (5y2 +y) + (5√2y + √2) = 0
    y (5y + 1) + √2 (5y + 1) = 0
    So, y = -1/5 (-0.2), -√2 (-1.4)
  10. I. 3x2– (9 + √3)x + 3√3 = 0,
    II. 3y2– (3 + 3√3)y + 3√3 = 0
    A) x > y
    B) x< y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relationship cannot be determined

    View Answer
    Option E
    Solution: 

    3x2– (9 + √3)x + 3√3 = 0
    (3x2 – 9x) – (√3x – 3√3) = 0
    3x (x – 3) – √3 (x – 3) = 0,
    So x = 3, √3/3 (0.58)
    3y2– (3 + 3√3)y + 3√3 = 0
    (3y2 – 3y) – (3√3y – 3√3) = 0
    3y (y – 1) – 3√3 (y – 1) = 0
    So x = 1, √3 (1.73)

 

 

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