Solve the System of Equations: Square Roots and xy=9

Question

Solve the following system of equations:

{x+y=61+6xy=9 \begin{cases} \sqrt{x}+\sqrt{y}=\sqrt{\sqrt{61}+6} \\ xy=9 \end{cases}

Video Solution

Solution Steps

00:00 Solve the following problem
00:03 Raise to the square
00:13 The square cancels the root
00:18 Use the shortened multiplication formulas to open the parentheses
00:30 The square cancels the root
00:43 Take the root of the first equation
00:48 Insert into our equation
01:05 Reduce wherever possible
01:16 Isolate X
01:19 This is the expression for X in terms of Y
01:22 Insert into our first equation
01:38 Open the parentheses properly, multiply by each factor
01:52 Arrange the equation so the right side equals 0
01:59 Use the root formula to find the solutions for Y
02:21 These are the possible solutions for Y
02:41 Now substitute them to find the possible solutions for X
03:16 This is the solution

Step-by-Step Solution

To solve this problem, we will follow these steps:

  • Step 1: Identify the equations and express one variable in terms of the other.
  • Step 2: Substitute into the other equation and simplify.
  • Step 3: Perform calculations to solve for the variable.
  • Step 4: Use the solution to find the second variable.

Let's work through the solution together:

Step 1: Given xy=9 xy = 9 , express y y as 9x \frac{9}{x} .

Step 2: Substitute into the first equation:

x+9x=61+6 \sqrt{x} + \sqrt{\frac{9}{x}} = \sqrt{\sqrt{61} + 6} .

Step 3: Simplify this equation. Let a=x a = \sqrt{x} and b=y b = \sqrt{y} .

Then, a+b=61+6 a + b = \sqrt{\sqrt{61} + 6} and ab=9=3 ab = \sqrt{9} = 3 .

Squaring both sides of the linear equation:

(a+b)2=61+6 (a + b)^2 = \sqrt{61} + 6 .

a2+2ab+b2=61+6 a^2 + 2ab + b^2 = \sqrt{61} + 6 .

Using ab=3 ab = 3 , we get 2ab=6 2ab = 6 .

This leads to a2+b2=61 a^2 + b^2 = \sqrt{61} .

Replacing a=x a = \sqrt{x} and b=y b = \sqrt{y} :

Let a2=x a^2 = x and b2=y b^2 = y and use the identity (ab)2=a2+b22ab=616(a - b)^2 = a^2 + b^2 - 2ab = \sqrt{61} - 6.

So, ab=616 a - b = \sqrt{\sqrt{61} - 6} .

Now let S=a+b S = a + b and P=ab P = ab from previous steps.

From S=61+6 S = \sqrt{\sqrt{61} + 6} and P=3 P = 3 , solve: t2St+P=0 t^2 - St + P = 0 .

This quadratic in t t gives solutions t=S±S24P2 t = \frac{S \pm \sqrt{S^2 - 4P}}{2} .

The quadratic roots are a=61+62±52 a = \frac{\sqrt{61} + 6}{2} \pm \frac{5}{2} and b=61+6252 b = \frac{\sqrt{61} + 6}{2} \mp \frac{5}{2} .

Thus, x=a2=(612+2.5)2 x = a^2 = (\frac{\sqrt{61}}{2} + 2.5)^2 or (6122.5)2 (\frac{\sqrt{61}}{2} - 2.5)^2 .

Similarly for y y .

Therefore, the solutions are:

x=6122.5 x = \frac{\sqrt{61}}{2} - 2.5 , y=612+2.5 y = \frac{\sqrt{61}}{2} + 2.5

or

x=612+2.5 x = \frac{\sqrt{61}}{2} + 2.5 , y=6122.5 y = \frac{\sqrt{61}}{2} - 2.5 .

Answer

x=6122.5 x=\frac{\sqrt{61}}{2}-2.5

y=612+2.5 y=\frac{\sqrt{61}}{2}+2.5

or

x=612+2.5 x=\frac{\sqrt{61}}{2}+2.5

y=6122.5 y=\frac{\sqrt{61}}{2}-2.5