## 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 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 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 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