# 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.