Deriving Quadratic Formulations From Areas of The Square

Let the quote symbol (") indicate the squaring operation.
The first display in the grid to the right includes  the red perimeter of a square partitioned into four sub-areas by intersecting blue lines. The intersection separates the blue lines into two segments. The length of the longer segment is designated (a) and the length of the shorter segment is designated (b). Therefore the four sub-areas will be designated (a"), (b") and (ab) and the total area of the square bounded by red lines (a + b)" equals (a" +  b" + 2ab). This analysis of the area of a square gives us the quadratic formulation (a + b)" = (a" + 2ab + b"). Note areas designated 2ab comprise four yellow right triangles.

Click the NEXT AREA button. Now the four yellow right triangles are each nested in the four corners of the red lined square. Note that the areas of the yellow triangles do not overlap, and the longest sides of the triangles form the perimeter of an inner square. Since the areas of the four triangles equal 2ab, the area of the inner square must equal the total area of the red lined square minus 2ab or (a + b)" - 2ab. In expanded terms (a" + b" + 2ab -2ab) = (a" + b"). If we designate the lengths of the long sides of the right triangle (c) then we have (a" + b") = (c"). This formulation is the pythagorean theorem for right triangles.

Click the NEXT AREA button. Now the four yellow right triangles have been rotated into the (c") area in such a manner that there areas do not overlap. Note that the lengths of the sides of the inner most square equal the length (a) of the longest leg of the right triangles minus the length (b) of the shortest leg of the right triangles, (a - b). Since the areas of the four triangles equal 2ab, the area of the inner most square must equal the area of the middle inner square (a" + b") - 2ab. So we have the quadratic formulation (a - b)" = a" - 2ab + b".  Click To Explore More About Quadratics Areas.

• The Application Of Areas To Parabolic Equations

  • Select an equation configuration, such as STF
  • Press NEXT AREA.
  • The first frame displays blue intersecting lines that bound B".
  • Press NEXT AREA.
  • The second frame displays blue and red intersecting lines that bound the areas associated with (b"), 2C1 and (a"), C1=b*a.
  • Press NEXT AREA.
  • The third frame displays red intersecting lines, yellow intersecting lines that bisect B" and blue and green intersecting lines that bound the areas associated with (f"), 2C2 and (e"), C2=f*e.
  • Press NEXT AREA.
  • The fourth frame displays all intersecting lines without labels.
  • Select and explore other equation configuration.

Exploring the Parabola 
      on  the Pythabacus

STEP1: Left Click Application. Press Enter/Return.

STEP2:Counting From the Rightmost Gold Bead, Left Click a Gold Bead at a position equal any number. Press Right Arrow (PRA). This step generates (B), which equals the displaced gold beads. The blue and lower right corner lines bound the area B"
STEP3: Left click another  gold  beads in or out of the displaced group (B). PRA. This step generates (a), the bead clicked in step 3 plus the total number of gold beads to its right, while (b) equals (B) - (a). If (B)>(a), the quadrilateral of brown beads above and to the left of of the a-group equals the product of (a) times (b), equals (C1) in the equation X" - BX + C1 = Y and C1 is positive. If (B)<(a) the quadrilateral of brown beads above and to the right of the displaced group plus the square of the number of gold beads in the displaced group equals (C1) and (C1) is negative.  The red, blue and lower right corner lines bound the areas associated with (b"), 2C1 and (a").
STEP4: Left click another  gold bead in any group. PRA. This step generates (e), the bead clicked in step 4 plus the total number of gold beads to its right, while (f) equals (B)-(e). If (B)>(e), the quadrilateral of brown beads above and to the left of of the a-group equals the product of (e) times (f), equals (C2) in the equation X" - BX + C2 = 0 and C2 is positive. If (B)<(e) the quadrilateral of brown beads above and to the right of the displaced group plus the square of the number of gold beads in the displaced group equals (C2) and (C2) is negative. The green, blue and lower right corner lines bound the areas associated with (f"), 2C2 and (e").
Press down arrow key to see step by step solution to the equation. For more details about generating various configurations of parabolic equations on the pythabacus click here.

To Graph a Parabola From the Areas of an Equation

• Select the  YAxis and go to the third frame again.
• If the sum of the segments or parts of the green line is equal the length of the blue line, then the roots will equal the lengths of the segments of the green line and the values of both roots are positive.
• The vertex is always equal one-half the length of the blue line (2.5)
• In this case the sum of the segments or parts of the green line is greater than the length of the blue line, and the first root is equal the length of the segment by which the sum of the green line is greater then the blue and its value is negative (-2), and the second root is equal the sum of the lengths of the segments of the green line and its value is positive (7).
• As noted above, the y-coordinate (K) of the vertex equals the sum of the areas with shared corners at the intersection of the green lines if one area equals h" a quadrant of B" with a corner at the intersection of yellow lines.
•  Or as in this case the y-coordinate (K) of the vertex equals the area of the square with a shared corner at the intersection of green lines and its opposite corner at the intersection of yellow lines (-20.25).
• The K value always has a negative value for these examples.
• In this case the area that elbows around (h") equals the Y-intercept (-14).
• Enter the equation given for the area represented above X"-5X-36=-22 or X"-5X-14=0 into DESMOS Graph Calclator.
• Verify varaible solutions predicted by area inspection and calculations above
• Explore other examples with the pythabacus and application of areas above, then see graph in the cartesian plane.

Identifying the Algebraic Elements of Completing the Square
by the Application of Areas

• Click application and press enter.

• Counting From the Rightmost Gold Bead, Left Click a Gold Bead at a position equal any Number. Press Right Arrow (PRA). This step generates (B), which equals the displaced gold beads. Let B" designate B-squared.

• Left click another  gold  beads in or out of the displaced group (B). PRA. This step generates (a), the bead clicked  plus the total number of gold beads to its right.

• Left click another  gold bead in any group. PRA. This step generates (e), the bead clicked in step 4 plus the total number of gold beads to its right, while (f) equals (B)-(e)

•  C1 the third term of the equation will equal the areas that elbow around an area equal a quadrant of B". These areas will share corners at the intersection of the red lines
• Or C1 will equal areas of a quadrant of B"that elbows around the area of the square with a shared corner at the intersection of red lines and its the opposite corner at the intersection of yellow lines.
• Below you will Press Down Arrow (PDA) to enter numbers in equations.
• And you will Press Up Arrow (PUA) to enter variables.
• PDA twice.
• C2 the third term of the equation equal (0) will equal the areas  that elbow around an area equal a quadrant of B". These areas will share corners at the intersection of the green lines
• Or C2 will equal areas of a quadrant of B"that elbows around the area of the square with a shared corner at the intersection of the green lines and its opposite corner at the intersection of the yellow lines.

• In the next equation, after X"-BX+/-C2=0, h" equals a quadrant of B" and h"-C2 equals (-K).
• If  C2 is negative,  the y-coordinate (K) of the vertex equals the sum of the areas with shared corners at the intersection of the green lines if one area equals h" a quadrant of B" with a corner at the intersection of yellow lines.
•  If C2 is positive the y-coordinate (K) of the vertex equals the area of the square with a shared corner at the intersection of green lines and its opposite corner at the intersection of yellow lines.
• PDA twice.
• (X-h) equals the square root of (-K) where X is the length of a segment of the intersecting green lines and h is a length of a side of a quadrant of B"
• Since K always has a negative value in these examples, (-K) kas a positive value.
• Remember the X that equals the length of the segment by which a green line  exceeds the length of the blue line has a negative value, otherwise the value of X is positive, while the value of h  is always positive.