## Quadratic Equations – Part 1

Type 1: $x^{2}+5x+6=&space;0$,  Find the roots for the equation?

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 the 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 change 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

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 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

## 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 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 $X>Y$