Quadratic Equation – Part 6

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}

Finally, the answer is y>x.

Quadratic Equations – Part 5

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 2x^{2}+x-206y^{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 312x^{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

Quadratic Equations – Part 3

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 \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} , \frac{8}{7}

Step 7: Change the sign -\frac{7}{6} , -\frac{8}{7}

 

Observe the diagram for the given equation

 

Quadratic Equations- Part 2

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

Quadratic Equations – Part 1

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