Geometric proof for x2 + 10 x = 39

Geometric proof for 3x=6

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Using Manipulatives

Start with a rectangle with dimension 3 and x

The equation tell us that the area of the rectangle is equal to 6.

We find the missing dimension ‘x’ by dividing 6 by 3. 6/3=2

X=2.

X2

Start with a square with side length x

X

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X2

5x/2

5x/2 5x/2

5x/2

Step 2.Then add the 10x.This is done by adding four rectangles with dimensions ‘x’ by ‘10/4’ to the four sides of the square. (10/4 simplifies to 5/2)

X

52

X2

5x/2x

5x/2 5x/2

5x/2

Step 3To complete the square, add four yellow squares to the corners.

X2

5x/2

5x/2 5x/2

5x/2

254

Each yellow square has dimensions of 5/2

52

Thus, the area of each yellow square is 25/4.

254

254

254

The blue square has an area of x2.

The four red rectangles have an area of 10x. [(5x/2) X 4]The four yellow squares have an area of 25. [(5/2) X 4]

X2

5x/2

5x/2 5x/2

5x/2

254

254

254

254

39

The equation tells us that x2 +10x is equal to 39. So, we can replace the blue and red shapes with a purple shape that has an area of 39.

254

254

254

254

39

To get the area of the entire shape, we add the area of the four yellow squares which is 25 to the area of purple shape which is 39 and get an area of 64 which is represented by the orange square.

64

This orange square has dimensions of 8.

8

To find x, we subtract 2(5/2) from 8 and we get 3.

64 8

52

52

3

X=3

Solve x2 -6x+9=0 using manipulatives Green=positiveRed=negative

x2

11

111

111

1

x x x

x

x

x

1.Start with big green square (x2).

2. Put nine small green square in the bottom right corner

3. Complete the square using six red rectangles

4. The dimensions of the large shape gives the factors for x2 -6x+9=0.The dimensions are x-3 and x-3

5.Solve for x.

X-3=0 X-3=0X=3 X=3

x

x

-1 -1-1

-1

-1

-1