Basic Engineering Mathematics, Fifth Edition

160 Basic Engineering Mathematics

(d) Comparing y=−^2 x^2 +^3 x+6(3)

with 0 =− 2 x^2 +x+5(4)

shows that, if 2x+1 is added to both sides

of equation (4), the RHS of both equations

will be the same. Hence, equation (4) may be

written as 2x+ 1 =− 2 x^2 + 3 x+6. The solu-

tion of this equation is found from the points

of intersection of the line y= 2 x+1andthe

parabola y=− 2 x^2 + 3 x+6; i.e., points G

and H in Figure 19.12. Hence, the roots of

− 2 x^2 +x+ 5 =0arex=− 1. 35 andx= 1. 85

Now try the following Practice Exercise

PracticeExercise 73 Solving quadratic

equations graphically (answers on page 348)

1. Sketch the following graphs and state the
   nature and co-ordinates of their respective
   turning points.
   (a) y= 4 x^2 (b) y= 2 x^2 − 1
   (c) y=−x^2 +3(d)y=−

1

2

x^2 − 1

Solve graphically the quadratic equations in prob-

lems 2 to 5 by plottingthecurves between thegiven

limits. Give answers correct to 1 decimal place.

2. 4x^2 −x− 1 = 0 ; x=−1tox= 1


3. x^2 − 3 x= 27 ; x=−5tox= 8


4. 2x^2 − 6 x− 9 = 0 ; x=−2tox= 5


5. 2x( 5 x− 2 )= 39. 6 ; x=−2tox= 3


6. Solve the quadratic equation
   2 x^2 + 7 x+ 6 =0 graphically, given that
   the solutions lie in the range x=−3to
   x=1. Determine also the nature and
   co-ordinates of its turning point.


7. Solve graphically the quadratic equation
   10 x^2 − 9 x− 11. 2 =0,giventhattherootslie
   betweenx=−1andx=2.


8. Plot a graph ofy= 3 x^2 and hence solve the
   following equations.
   (a) 3x^2 − 8 =0(b)3x^2 − 2 x− 1 = 0


9. Plot the graphsy= 2 x^2 andy= 3 − 4 xon
   the same axes and find the co-ordinates of the
   points of intersection. Hence, determine the
   roots of the equation 2x^2 + 4 x− 3 =0.
   10. Plot a graph of y= 10 x^2 − 13 x−30 for
   values of x between x=−2andx=3.
   Solve the equation 10x^2 − 13 x− 30 =0and
   from the graph determine
   (a) the value ofywhenxis 1.3,
   (b) the value ofxwhenyis 10,
   (c) the roots of the equation
   10 x^2 − 15 x− 18 =0.

19.3 Graphical solution of linear

and quadratic equations

simultaneously

The solutionoflinear and quadratic equations simul-

taneouslymay be achieved graphically by

(a) plottingthe straight line and parabola on the same

axes, and

(b) noting the points of intersection.

The co-ordinates of the points of intersection give the

required solutions.

Here is a worked problem to demonstrate the simulta-

neous solution of a linear and quadratic equation.

Problem 7. Determine graphically the values of

xandywhich simultaneously satisfy the equations

y= 2 x^2 − 3 x−4andy= 2 − 4 x

y= 2 x^2 − 3 x−4 is a parabola and a table of values is

drawn up as shown below.

x − 2 − 1 0 1 2 3

y 10 1 − 4 − 5 − 2 5

y= 2 − 4 xis a straight line and only three co-ordinates

need be calculated:

x 0 1 2

y 2 − 2 − 6

The two graphs are plotted in Figure 19.13 and the

points of intersection, shown as AandB, are at co-

ordinates(− 2 , 10 )and( 1. 5 ,− 4 ). Hence, the simul-

taneous solutions occur when x=− 2 , y= 10 and

whenx= 1. 5 ,y=− 4.

These solutions may be checked by substituting into

each of the original equations.