# Thread: Solve by completing the square

1. ## Solve by completing the square

Hello! I would like some help with this task.

The perimeter of a square and a rectangle are equal. The length of the rectangle is 11 cm and the area of the square is 4 cm ^ 2 more than the area of the rectangle. Find the side of the square.

Thank you!

2. Originally Posted by Anna55
Hello! I would like some help with this task.

The perimeter of a square and a rectangle are equal. The length of the rectangle is 11 cm and the area of the square is 4 cm ^ 2 more than the area of the rectangle. Find the side of the square.

Thank you!

Let the side of the square be "x".

Let the width of the rectangle be "y".

The perimeter of the square is $\displaystyle 4x$

The perimeter of the rectangle is $\displaystyle 22+2y$

The perimeters are equal, so $\displaystyle 4x=22+2y$

The square area is $\displaystyle x^2$

The rectangle area is $\displaystyle 11y$

$\displaystyle x^2=11y+4$

Then you can eliminate y and solve for x in the simultaneous equations..
or, as your title suggests, substituting for y and completing the square.

$\displaystyle 4x=22+2y$

$\displaystyle x^2=11y+4$

3. Thank you for your help!
I tried to solve the two equations, however I got the wrong answer. How should I do if i want to solve it by completing the square?

4. The square is on the "x" term, so we want to form a quadratic equation with "x".

This means we need a substitution for y, so we use the linear equation to do that.

$\displaystyle 4x=22+2y\Rightarrow\ 2x=11+y\Rightarrow\ y=2x-11$

Use this y in the 2nd equation.

$\displaystyle x^2=11y+4\Rightarrow\ x^2=11(2x-11)+4$

$\displaystyle x^2=22x-121+4\Rightarrow\ x^2-(22x-121)=4\Rightarrow\ x^2-11x-11x+121=4$

$\displaystyle \Rightarrow(x-11)^2=4$

Take positive and negative square roots...

$\displaystyle x-11=\pm2$

$\displaystyle x=11\pm2$

5. I cleaned that up.

6. Thank you very much Archie Meade! It helped a lot and I understand everything now.