7.5 Graphical method of solving a
Quadratic equations:
We have learnt how to draw graph for an equation of
type y=mx+c(Where m is a constant).
We have also observed that the equation of this
type represents a straight line.
Let us learn to solve quadratic equation of type ax^{2}
+bx+ c =0
We can solve the equation by two methods
First method:
ax^{2} +bx+ c = 0
can be written as
ax^{2}= bxc
Let each be equal to y
So we have two equations y = ax^{2 }and y
=bxc
Draw the graph for both these equations. The
intersecting points of the two graphs are solutions to the given equation ax^{2}
+bx+ c=0
Note that y = =bxc is an
equation to a line. Problem 7.5.1 illustrates this method.
Second method:
Draw the graph ax^{2} +bx+
c and then find the points on the graph which touches xaxis (I.e. when
y=0).The xcoordinates of the points on the graph, whose ycoordinates are
zero, are roots of the
given equation. Problem 7.5.2 illustrates this method.
7.5 Problem 1: Draw the graph for
y=2x^{2} and y= 3+x and hence
1. Solve the equation 2x^{2}x3=0
2. Find the value of _{}
Solution:
1. Solving of equation 2x^{2}x3=0
Step
1: For few values of x tabulate the values of y (=2x^{2})
as shown below:
Step
2: On a graph sheet mark the (x, y) coordinates. Join these points by a
smooth curve. This smooth curve is called ‘parabola’ Step
3: For two values of x tabulate the values of y
(=3+x) as shown below
Why did
we ask you to find coordinates for only 2 values of x? (y=3+x is of type y=mx+c
and it represents a straight line. To draw a straight line, two points are
enough) Step
4: On a graph sheet mark these two (x,y)
coordinates. Join these two points. The
parabola and the straight line cut each other at two points. They are (1, 2)
and (1.5, 4.5). Their
x coordinates are 1 and 1.5 respectively. 1
and 3/2 satisfy the given equation 2x^{2}x3=0. Verification: The given equation 2x^{2}x3=0
is of the form ax^{2} +bx+ c =0. We have
learnt that the roots of this equation are _{}x = [b _{}_{} (b^{2}4ac)]/2a Here
a=2, b= 1,c= 3 _{}_{} (b^{2}4ac) = _{} (1+24)= _{}5 The
roots are (1_{}5)/4 = 1 and 3/2 which are as derived using the graphical
method 


2. Finding value of _{}: We
need to find the value of y when x =_{}. The graph for parabola is of the form y=2x^{2}.
Thus if x =_{} then y=2x^{2}=
6 So we need to find the value of x when y=6. From the graph we can see that
there are two values of x (circled points in the
graph in I Quadrant and II Quadrant) which satisfy y=6 We notice that x coordinate of the parabola at y=6 are 1.7 and +1.7 which gives the value of _{} 
Exercise and observations:
1. Draw few graphs for the equation y =mx^{2}
for few values of m (both +ve and –ve) and observe the following:

All graphs are parabolas and pass through the origin

They are all symmetric about yaxis.
2. Draw few graphs for the equation x =my^{2}
for few values of m (both +ve and –ve) and observe the following:

All graphs are parabolas and pass through the origin

They are all symmetric about x axis.
7.5 Problem 2: Solve the equation 2x^{2}+3x5=0
Solution:
Step
1: For few values of x tabulate the values of y(=2x^{2}+3x5) as
given below:
Step
2: On a graph sheet mark these (x, y) coordinates. Join
these points by a smooth curve. This smooth curve is a parabola. We
need to find the point on graph when 2x^{2}+3x5=0( i,e when y=0) We
notice that the graph touches the x axis (note that y=0 for any point on x
axis) at x=
2.5(= 5/2) and at x=1. Therefore
1 and 5/2 are the roots of the given equation. Verification: The given equation
2x^{2}+3x5=0 is of the form ax^{2} +bx+
c =0 We have learnt that the roots of this
equation are _{}x = [b _{}_{} (b^{2}4ac)]/2a Here
a=2, b= 3,c= 5 _{} Determinant b^{2}4ac = 9+40=49 _{}_{} (b^{2}4ac) = _{}7 The
roots are (3_{}7)/4 I.e.
= 1 and 5/2 are the roots which we derived using the graphical method 

Exercise and observations:
Draw graphs for the following equations in both the
methods and observe the following

Equation

Method1 
Method 2 
Reason 



Draw 2 graphs 
Observations 
Draw 1 graph for 
Observations 
Determinant = b^{2}4ac = 
1 
2x^{2}+2x15=0 
y = 2x^{2} y
= 2x+15 
The
graphs meet at two points:(5,0),(3,0) 
2x^{2}+2x15 
The
graph touches x axis at 2 points and thus has two roots. 
460=64 = 8^{2} (perfect
square) 
2 
4x^{2}4x+1=0 
y = 4x^{2} y
= 4x1 
The
graphs meet at one point: (1/2,0) 
4x^{2}4x+1=0 
The
graph touches x axis at 1 point and
has only one root 
1616=0 (zero) 
3 
x^{2}6x+10=0 
y = x^{2} y
= 6x10 
The
graphs do not meet at all! 
x^{2}6x+10=0 
The
graph does not touch x axis and thus no roots. 
3640
= 4 (negative) 
7.5 Summary of learning
No 
Points studied 
1 
Quadratic
equations can be solved by drawing graphs 