Problem Statement: $x^{2}-7\sqrt{3}x+36$ ; $y^{2}-12\sqrt{2}y+70$ which is greater x or y?

Solution: When you find any root digit in the equation it is difficult to solve such a problem. So here is the shortcut method is

$x^{2}-7\sqrt{3}x+36$

Step 1: Just divide the constant  36 with the middle number which in the root

$\frac{36}{3}=12$

Step 2: Forgot,$^{\sqrt{3}}$ now the equation is

$x^{2}-7x+12$

Step 3: Follow the procedure which is shown in part 1

Observe the diagram as shown below

Step 4: Finally add $^{\sqrt{3}}$ to the resultant roots

$3\sqrt{3},4\sqrt{3}$

$y^{2}-12\sqrt{2}y+70$

Step 1: Just divide the constant  70 with the middle number which is the root

$\frac{70}{2}=35$

Step 2: Forgot, $^{\sqrt{2}}$ now the equation is

$y^{2}-12y+35$

Step 3: Follow the procedure which is shown in part 1

Observe the diagram as shown below

Step 4: Finally add $^{\sqrt{2}}$ to the resultant roots

$7\sqrt{2},5\sqrt{2}$

Type: Inequalities

When we got such cases like below in the given problem the answer is cannot be determined.

• (-,-) and (+,-)
• (-,-) and (-,-)
• (+,-) and (+,-)

Example 1: $x^{2}+x-20$ ; $y^{2}-y-30$ which is a greater value x or y from both equations?

It is matched with the first case so the answer is cannot be determined or relationship does not exist.

Example 2$x^{2}+x-20$$6y^{2}+y-1$ which is a greater value x or y from both equations?

It is matched with the third case so the answer is cannot be determined or relationship does not exist.

Example 3$12x^{2}-2x-4$ $y^{2}-y-30$

It is matched with the second case so the answer is cannot be determined or relationship does not exist.

Example 4: If you find  (-, +)  and (+,+) pair in the given problem the answer is (-,+)  >  (+,+)

Example: $y^{2}+8y+15$ $x^{2}-8x+15$

The answer is $X>Y$

Type – Inequalities

Problem 1: $y^{2}=9$$x=\sqrt{9}$ which is greater x or y?

Solution : $y^{2}=9$          $x=\sqrt{9}$

$y=\pm&space;3$        $x=3$

$y\leq&space;x$

Problem 2: $y^{2}=64$  ,$x=\sqrt{64}$ which is greater x or y?

Solution: $y^{2}=64$         $x=\sqrt{64}$

$y=\pm&space;4$         $x=4$

$y\leq&space;x$

Type 3:  $42x^{2}+97x+56$

Basic method:

To solve this problem in basic method we can use one formula

i.e.  $\frac{-b\pm&space;\sqrt{b^{2}-4ac}}{2a}$    here a=42    b=97     c=56

By using this formula we can not solve this problem in less time. So to solve this problem in less time there an alternative method as shown below.

Shortcut Method:

Step 1: Find the factors for first term 42

The factors are 2,3,6,7,42,1

6,7 are the appropriate factors for this case (See the below diagram for more details)

Step 2: Find the factors for constant 56

The factors are 2,3,7,8,56,1

7,8 factors are the appropriate factors for this case (See the below diagram for more details)

Step 3: Select 7 in the first one and 7 in the second one and multiply both

7*7=49

Step 4: Select 6 in the first one and 8 in the second one and multiply both

6*8=48

Step 5: Add both 49+48=97 which is equal to middle number i.e.97

Step 6: Divide the,$\frac{49}{42}$ $\frac{48}{42}$

The result after the division –   $\frac{7}{6}&space;,&space;\frac{8}{7}$

Step 7: Change the sign $-\frac{7}{6}&space;,&space;-\frac{8}{7}$

Observe the diagram for the given equation

Type 2:   $2X^{2}+8X+8$

Basic method: $2X^{2}+8X+8$

$2x^{2}+4x+4x+8$

$2x(x+2)+4(x+2)$

$(2x+4)(x+2)$

$2x=-4$            $x=-2$

$x=-\frac{4}{2}$

$x=-2$

Shortcut method:

• Step 1: Multiply the first number “2” and the constant 8
• 2*8=16
• Step 2: Find the factors for 16
• 2,4,8,16,1 are the factors.
• Step 3: Select the factors such that when we add we should get the middle number (8 in the given equation) and when we multiply we should get constant (16).
• 4 and 4 are such factors which will give sum as 8 and multiplication as 16.
• If we take 1 and 6, multiplication is 6  but the sum is 7 which is not equal to 5.
• So the appropriate factors for the given equation are 3 and 2.
• Step 4: Change the sign for the  factors
• Since it is ”+8x” in the given equation we need to change the sign for the factors  i.e. -4,-4
• If it is “-8x” in the equation we need to change the sign for the factor as 4, 4.
• Step 4: Divide the resultant factors by the first number 2
• $-\frac{4}{2},-\frac{4}{2}$
• $-2,-2$  are the final roots.

Observe the diagram below

Type 1:  $X^{2}+5X+6$

Basic method: $x^{2}+5x+6$

$x^{2}+3x+2x+6$

$x(x+3)+2(x+3)$

$(x+2)(x+3)$

$x+2=0$            $x+3=0$

$x=-2$                $x=-3$

Short Cut Method:

• Step1: Find the factors for constant (In the given equation it is 6)
• 2,3,6,1 are the factors for 6
• Step 2: Select the factors such that when we  add we should get middle number (5 in the given equation) and when we multiply we should get constant (6).
• 2 and 3 are such factors which will give sum as 5 and multiplication as 6.
• If we take 1 and 6, multiplication is 6  but sum is 7 which is not equal to 5.
• So the appropriate factors for the given equation are 3 and 2.
• Step 3: Change the sign  for the  factors
• Since it is ”+5x” in the given equation we need to chance the sign for the factors  i.e. -3,-2
• If it is “-5x” in the equation we need to change the sign for the factor as 3, 2.

Note: 2 and 3 are co-primes. So we should select co-prime factors.

observe the below diagram

Simplifications

Problem 1: Simplify (33076161)^⅓ + (279841)^¼ + (243)^⅕ = ?

(33076161)^1/3 + (279841)^1/4 + (243)^1/5 = ?

Given options are

• 344
• 345
• 346
• 347
• None of these

The solution involves four steps

• Finding the cube root of the number 33076161
• Finding the fourth root of 279841
• Find the fifth root of 243
• Adding all the three numbers
1. Finding the cube root of the number 33076161

TP SP FP – We will be finding the numbers in these three places as below.

• Divide the number into three parts
• First number (FN) : 33
• Second number (SN): 076
• The third number (TN): 161

To find the cube root, we will use TN, then FN and finally SN.

• TN ends with 1, so the number in the one’s place (FP) is 1. (Refer Table 1 below)

TP SP 1

• FN is 33, which in between the cube of 3 and 4, so we have to choose the small number which is 3 will be in third place (TP). (Refer table 1 below)

3 SP 1

• To find the number in second place, we can use the formula – 3*(TP^2)*SP= second digit of (TN-FP)
• 3*(1^2)*SP= Second digit of (161-1) = Second digit of 160 = 6
• 3*(1^2)*SP=6
• SP=6/3=2

3 2 1

• Finally, the cube root of (33076161)⅓ is 321

2. Finding the fourth root of 279841

• First, find the square root of the number 279841
• The square root of 279841 is 529
• Again find the square root of 529
• The square root of 529 is 23
• (23^4)¼=23

3. Find the fifth root of 243

• Rewrite 243 as 3^5
• Cancel the common factor of 5
• And the answer is 3

4. Then, the final step is adding all the three numbers 321+23+3=347

Table 1: Square, Cube, Numbers Ranging 0 – 10

 Number x Square x2 Cube x3 1 1 1 2 4 8 3 9 27 4 16 64 5 25 125 6 36 216 7 49 343 8 64 512 9 81 729 10 100 1000

Problem2:  55555+5555+555+55+5+=?

To find the solution in an easy way first we have to count the no.of digits in a number and we have to choose the highest digit number

Step1: In this number 55555, the no.of digits are 5

5555=4

555=3

55=2

5=1

Step 2: Multiply as shown in the below

5*5=25 here 5 is the number and 2 is carried

5*4=20 add the previous carry to the number 20 then 20+2=22 2 is the number and 2 is carried

5*3=15 add the previous carry to the number 15 then 15+2=17 7 is the number and 1 is carried

5*2=10 add the previous carry to the number 10 then 10+1=11 1 is the number and 1 is carried

5*1=5 add the previous carry to the number 10 then 5+1=6 so 6 is the number

And finally, the answer is 61725.

 Number digit* no.of digits sum Carry Forward (CF) sum+CF Final Digit 55555 5*5 25 0 25 5 5555 5*4 20 2 22 2 555 5*3 15 2 17 7 55 5*2 10 1 11 1 5 5*1 5 1 6 6

Final Digit = second digit of (sum+CF)

Carry Forward = first digit of previous (sum+CF)

So the Final sum is 61725

Problem 3: 0.4444+0.444+0.44+0.4=?

To find the solution for this there is a shortcut which is similar to the above problem

Here we have to choose the smallest digit number

The smallest digit is 0.4 the no.of digits are 1

0.44=2

0.444=3

0.4444=4

Step 2: multiply as shown in the below

4*1=4

4*2=8

4*3=12 here 2 is the number and 1 is carried

4*4=16 add the previous carry to the number 16 then 16+1=17

And finally, we get 17284

After four digits there is a decimal point so the answer is 1.7284

 Number digit* no.of digits sum Carry Forward (CF) sum+CF Final Digit 0.4 4*1 4 0 4 4 0.44 4*2 8 0 8 8 0.444 4*3 12 1 2 2 0.4444 4*4 16 1 17 17

Final Digit = Second digit of (sum+CF)

Carry Forward = First digit of previous (sum+CF)

So the Final sum is 1.7284

Problem 4: 3333.666+33.6666+333.66+3.6=?

This can be divided into two parts

(3333+33+333+3)+(0.666+0.6666+0.66+0.6)

4            2      3      1          3             4             2        1

Part-1 and 2 can solve by using the above method

Choose highest digit number choose the lowest digit number

3*4=12 2 is number 1 is carry 6*1=6

3*3=9 9+1=10 0 is number 1 is carry 6*2=12 2 is number and 1 is carry

3*2=6 6+1=7 6*3=18 18+1=19 9 is the number 1 is carry

3*1=3 6*4=24 24+1=25

Ans is 3702

ans is 2.5926

3702+2.5926=3704.5926

Trick: 61+16

When it is addition we have to multiply with 11

11*(1+6) or 11*(6+1)

11*7=77

Trick: 76-67

When it is subtraction we have to multiply with 9

9(7-6)

9*1=9

Problem: 53+35-56-65+63+36-98-89=?

53+35-(56+65)+63+36-(98+89)

11(8-11+9-17)

11*-11

-121

Problem: 53-35+83-38+68-86=?

9(2+5-2)

9*5=45

Trick: 65*75

1.7+1=8

2.6*8=48

3.difference between 6 and 7 is 1

4.last digit is 25

5.4825+1*50

=4825+50

=4875

Example: 25*95

9+1=10

2*10=20

2025

Difference between 2 and 9 is 7

2025+7*50

2025+350

2375