## Quadratic Equation – Part 9

Type 9: $x^{2}-5x-336$

Finding roots for the above problem is difficult. So here is the easy method to solve such kind of problems.

Step 1: Find the LCM for the constant number 336

Observe the below diagram

Step 2: Use the formula – Multiply all even factors and all odd factors separately.

Here the even factor is 2 and the odd factor is 7 and 3

(2*2*2*2), (7*3)

16, -21 (When we add these two we will get “-5”)

Step 3: When we add these two factors we will get -5 and when we multiply, we get 336

16-21=-5           16*21=336

The roots are 16, -21.

Example 2:  $x^{2}-239x-972$

Step 1: Find the LCM for the constant number 336

Observe the below diagram

Step 2: Use the formula – Multiply all even factors and all odd factors separately

Here the even factor is 4 and the odd factor is 3

(4) ; (3*3*3*3*3)

4, -243

Step 3: When we add  these two factors we will get -239 and when we multiply we will get 972

4-243=-239; 4*243=239

The roots are 4, -243

## Quadratic Equations – Part 8

Type 8: $\sqrt{y}=\frac{(19^{\frac{3}{2}})}{y}$   ;     $\frac{15}{\sqrt{x}}-\frac{9}{\sqrt{x}}=\sqrt{x}$ Which is greater x or y?

Solution:     $\sqrt{y}=\frac{(19^{\frac{3}{2}})}{y}$

$y^{\frac{1}{2}}y=(19^{\frac{3}{2}})$

Remember this formula $a^{m}.a^{n}=a^{m+n}$

$y^{\frac{1}{2}+1}=(19)^{\frac{3}{2}}$

$y^{\frac{3}{3}}=19^{\frac{3}{2}}$

$y=19$

Solution:  $\frac{15}{\sqrt{x}}-\frac{9}{\sqrt{x}}=\sqrt{x}$

$\frac{15-9}{\sqrt{x}}=\sqrt{x}$

$x=6$

The final answer is $y>x$

## Quadratic Equations – Part 7

Type 7:  $2x^{2}-(4+\sqrt{13})x+2\sqrt{13}$

Basic method : $2x^{2}-4x-\sqrt{13}x+2\sqrt{13}$

$2x(x-2)-\sqrt{13}(x-2)$

$(2x-\sqrt{13})(x-2)$

$2x-\sqrt{13}=0$            $x-2=0$

$x=\frac{\sqrt{13}}{2}$                          $x=2$

Shortcut method:

Step 1: Consider the middle term  $(4+\sqrt{13})$

Step 2: Split the term into 4 and $\sqrt{13}$

Step 3: Divide  4 and $\sqrt{13}$  by the first term 2.

$\frac{4}{2},\frac{\sqrt{13}}{2}=2,\frac{\sqrt{13}}{2}$

The roots of the given equation are  $2,\frac{\sqrt{13}}{2}$

$10y^{2}-(18+5\sqrt{13})y+9\sqrt{13}$

Step 1: Consider the middle term  $(18+5\sqrt{13})$

Step 2: Split the term into 18 and $5\sqrt{13}$

Step 3: Divide  18 and $5\sqrt{13}$  by the first term 10

$\frac{18}{10},\frac{5\sqrt{13}}{10}=1.8,\frac{\sqrt{13}}{2}$

The roots of the given equation are  $1.8,\frac{\sqrt{13}}{2}$

The final answer is $x\geq&space;y$

## 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

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

## 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 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 $X>Y$

## Quadratic Equation – Part 4

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: $y^{2}=64$         $x=\sqrt{64}$

$y=\pm&space;4$         $x=4$

$y\leq&space;x$

## 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.  $\frac{-b\pm&space;\sqrt{b^{2}-4ac}}{2a}$    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.

## 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