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.

Leave a Reply

Your email address will not be published. Required fields are marked *