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 ax2 +bx+ c =0

 

We can solve the equation by two methods

 

First method:

ax2 +bx+ c = 0 can be written as

ax2= -bx-c

Let each be equal to y

So we have two equations y = ax2 and y =-bx-c

Draw the graph for both these equations. The intersecting points of the two graphs are solutions to the given equation ax2 +bx+ c=0

Note that y = =bx-c is an equation to a line. Problem 7.5.1 illustrates this method.

 

Second method:

Draw the graph ax2 +bx+ c and then find the points on the graph which touches x-axis (I.e. when y=0).The x-co-ordinates of the points on the graph, whose y-co-ordinates are zero, are roots of  the given equation. Problem 7.5.2 illustrates this method.

 

7.5 Problem 1:  Draw the graph for y=2x2 and y= 3+x and hence

 

1. Solve the equation 2x2-x-3=0

2. Find the value of

 

Solution:

1. Solving of equation 2x2-x-3=0

Step 1: For few values of x tabulate the values of y (=2x2) as shown below:

       

x à

0

1

-1

2

-2

y à

0

2

2

8

8

(x, y)

(0,0)

(1,2)

(-1,2)

(2,8)

(-2,8)

Step 2: On a graph sheet mark the (x, y) co-ordinates. 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

        

x à

-1

1

y à

2

4

(x, y)

(-1,2)

(1,4)

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) co-ordinates. 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 co-ordinates are -1 and 1.5 respectively.

-1 and 3/2 satisfy the given equation 2x2-x-3=0.

Verification:

The given equation 2x2-x-3=0 is of the form ax2 +bx+ c =0. We have learnt that the roots of this equation are         x = [-b  (b2-4ac)]/2a      Here a=2, b= -1,c= -3  (b2-4ac) =  (1+24)= 5

The roots are (15)/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=2x2. Thus if x =  then y=2x2= 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 co-ordinate 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 =mx2 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 y-axis.

2. Draw few graphs for the equation x =my2 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  2x2+3x-5=0

 

Solution:

Step 1: For few values of x tabulate the values of y(=2x2+3x-5) as given below:

         

x à

0

1

-1

2

-2

-3

y à

-5

0

-6

9

-3

4

(x, y)

(0,-5)

(1,0)

(-1,-6)

(2,9)

(-2,-3)

(-3,4)

 

Step 2: On a graph sheet mark these (x, y) co-ordinates.

Join these points by a smooth curve. This smooth curve is a parabola.

We need to find the point on graph when 2x2+3x-5=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 2x2+3x-5=0 is of the form ax2 +bx+ c =0

 We have learnt that the roots of this equation are

x = [-b  (b2-4ac)]/2a    Here a=2, b= 3,c= -5

 Determinant b2-4ac = 9+40=49  (b2-4ac) = 7

The roots are (-37)/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

= b2-4ac =

1

2x2+2x-15=0

y = 2x2

y = -2x+15

The graphs meet at two points:(-5,0),(3,0)

2x2+2x-15

The graph touches x axis at 2 points and thus has two roots.

4-60=64 = 82

(perfect square)

2

4x2-4x+1=0

y = 4x2

y = 4x-1

The graphs meet at one point: (1/2,0)

4x2-4x+1=0

The graph touches x axis at 1  point and has only one root

16-16=0

(zero)

3

x2-6x+10=0

y = x2

y = 6x-10

The graphs do not meet at all!

x2-6x+10=0

The graph does not touch x axis and thus no roots.

36-40 = -4

(negative)

 

 

 

7.5 Summary of learning

 

 

No

Points studied

1

Quadratic equations can be solved by drawing graphs