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.