Lecture1.8 Geometric method for solving or evaluating quadratic expressions
1. GEOMETRIC METHOD OF FINDING ROOTS OF / EVALUATING QUADRATIC.
Construction
Consider the roots of the quadratic equation for
simplicity we have taken the coefficient of as 1. Starting from the origin O
of the Cartesian coordinate system draw a line segment OB of 1 unit to the
right along X-axis. Then turn 90 anticlockwise and draw another line
segment BC of length b , this time parallel to the Y-axis. Again turn 90
O B
C
X
P
Q
D
1
b
c
Z
Geometric method of finding roots of a quadratic.
S
R
2. anticlockwise and draw a line segment CD parallel to negative direction of
X-axis. Join OD and let its midpoint be Z . Draw a circle with center at Z and
radius OZ and let it cut the line segment BC at P and Q. Now OP,PD,OQ and
QD.
Observation
It may be easily seen that ∠𝑂𝑃𝐷 and ∠𝑂𝑄𝐷 are right angles, being angles
subtended by semicircle OSD on the circumference OPQD.
The equation of the line segment is obviously 𝑥 = 𝑏 and the equation of the
line segment CD is𝑦 = 𝑐.
If the circle intersects CD at R, we have RC = AB = 1 unit and RD = 𝑐 − 1 units.
The coordinates of the point D are (−(𝑐 − 1),b) or, (1 − 𝑐, 𝑏); so that the
coordinates of Z, midpoint of OD are . Radius of the circle is the
distance OZ, given by
So the equation of the circle is given by
…….(A)
To find the coordinates of P and Q the intersection of the circle with the line
segment BC who has the equation 𝑥 = 1, ………………(B)
we have to solve equation (A) and equation (B).
Substituting 𝑥 = 1 in equation (A) we get,
Or,
3. We need not solve this equation; but merely compare this equation with the
equation ………………………(D)
It is easy to observe that sum of the roots of equation (C) is 𝑏, while that of
equation (D) is −𝑏, they are negatives of each other. But the product of roots
in both cases is 𝑐. So the roots of the two equations are negatives of each
other. If we denot 𝐏𝐁 = 𝑦 , 𝐐𝐁 = 𝑦 be the roots of the equation (C), then
the roots of the equation (D) must be −𝑦 , −𝑦 respectively.
Thus we get the roots of the quadratic equation by a geometrical construction.
Roots of
This is just a little generalization of the above result.
Just write in place of Now the
roots are not -PB and -QB, but respectively.
The roots are negative slopes of PO and QO instead of the segments.
Note: All these results are valid only if the roots are real!
Else the semicircle on the diameter OD shall not touch the line segment BC.
Note: When we take any arbitrary ray from the origin and bounce it
successively off the following perpendicular segments, the end point of the ray
may not hit the end point of the last segment. Naturally we have to adjust and
readjust the second rectangular path to hit the last point of the first
rectangular path.
Note: When this method is followed, it is not necessary to graph the quadratic
and find the x-intercepts to show the real roots.
Evaluation of
This is a generalization in another direction;
to evaluate
In the figure below CE = d ray if the ray OP or OQ shot from O bounces off the
line segment BC and meets CD at E . As before let m and n be the roots of
4. and the ray PD meet CE at D . What is c by the way? Not
to worry much. This c is product of roots of 𝑥 + 𝑏𝑥 + 𝑐 = 0 whose sum is –b
and that is just the equation we discussed just before. If denote its roots, we as
𝛼 and 𝛽 have 𝛼 + 𝛽 = −𝑏, 𝛼𝛽 = 𝑐
and for
all values of x.
Now .
So,
so also
O B
C
X
P
Q
D
1
b
c
Z
Geometric method of evaluating a quadratic.
S
R
E
5. But by remainder theorem, is remainder when is divided by
and is remainder when is divided by which is
in both cases.
Thus geometrically, DE = d-c is the remainder when is divided
by or , where as 𝛼 and 𝛽 are roots of
Is there anything special about or c ? Nothing except that 𝛽 are the
roots of So whatever may be 𝛽, if they are real
numbers, they must be roots of some for some c.
Then DE= d - c is the remainder left when is divided by
or . This is a geometrical meaning of d – c .
As a matter of fact, must be reduced to monic form (by
dividing throughout by ‘a’, coefficient of highest degree term. Obviously then
the segment DE shall not be remainder when is divided by
or but remainder divided by ‘a’.
A Deduction:Finding square root of a number ‘a’. Take the 𝑥 − 𝑎 = 0 and
convert it into 1𝑥 + 0𝑥 − 𝑎 = 0.
O A
1
B
C
X
Finding square root
a
√a
6. As before take a line segment of length 1 (coefficient of 𝑥 ) i.e. OA, turn 90
left and take a segment of 0 length, then turn again left and take a segment of
length – a to left (of length ‘a’ to the right of course i.e. AC). Take a ray from
O, bounce it off Y-axis and extend it to C. (if not, adjust and readjust it until the
second path hits the point C.). A well known fact from geometry is 𝐴𝐵 = 1. 𝑎
so that 𝐴𝐵 = √𝑎.
Another Deduction:
Finding complex roots.
For an animated visualization of complex roots visit
https://www.geogebra.org/m/U2HRUfDr#:~:text=Visualizing%20the%20re
al%20and%20complex,the%20graph%20as%20x%2Dintercepts.&text=This
%20visual%20imagines%20the%20cartesian,axis)%20of%20the%20comple
x%20plane.
X
Y
O
A
P
Q
R Complex Roots
S
T
U
V
𝑦 = 𝑥 + 𝑏𝑥 + 𝑐
C
B
W
Z
7. Let CAB be the parabola be the graph of quadratic equation
which has complex roots so that it does not cross the X-
axis. Needless to prove that its vertex A has coordinates, and
the equation of its axis is
To solve the quadratic equation 𝑦 = 𝑥 + 𝑏𝑥 + 𝑐 with complex roots we take
PO =1 unit, turn left and take OQ = b units again turn left and take QR = c
units. An orthogonal path from P to R is not possible since the equation has no
real roots. The real part of both roots shall still be −
𝑏
2
And the imaginary
parts shall be
To show this in the figure, we make some construction. Draw QU = OP = 1
unit. This is on extension of the line segment for c. Draw a circle taking
diameter UR and center at its midpoint. Let the circle meet the X-axis at S.
Now 𝑄𝑆 = 𝑄𝑈. 𝑄𝑅 , or in view of the fact that ∠QSR is a
right angle being the angle in the half circle and QS is perpendicular to UR.
Draw a circle STZ with radius QS Extending TV so that it cuts the circle
UPR at W, we get VW = - VT as QS and VT are perpendicular to each other ,
being parallel to coordinate axes and this perpendicular from center QV on
chord TV bisects it.. To determine this , note that QT is radius of the circle
STZW = QS = . So we have
Or, , so Note that this is
the discriminant of the quadratic and it is the square root of VA , distance of
the vertex of the parabola from the x-axis.
