FINDING NATURE OF QUADRATIC EQUATION BY GRAPHING

Finding Nature of Quadratic Equation by Graphing :

Here we are going to see some example problems of finding nature of solution of quadratic equation with graph.

To obtain the roots of the quadratic equation ax2 + bx + c = 0 graphically, we first draw the graph of y = ax2 +bx +c .

The solutions of the quadratic equation are the x coordinates of the points of intersection of the curve with X axis.

Finding the Nature of Solution of Quadratic Equations Graphically

To determine the nature of solutions of a quadratic equation, we can use the following procedure.

(i) If the graph of the given quadratic equation intersect the X axis at two distinct points, then the given equation has two real and unequal roots.

(ii) If the graph of the given quadratic equation touch the X axis at only one point, then the given equation has only one root which is same as saying two real and equal roots.

(iii) If the graph of the given equation does not intersect the X axis at any point then the given equation has no real root.

Finding Nature of Quadratic Equation by Graphing - Questions 

Question 1 :

Graph the following quadratic equations and state their nature of solutions.

(i)  x2 + x + 7 = 0

Solution :

Draw the graph for the function y = x2 + x + 7

Let us give some random values of x and find the values of y.

x

-4

-3

-2

-1

0

1

2

3

4

x2

16

9

4

1

0

1

4

9

16

x

-4

-3

-2

-1

0

1

2

3

4

+7

7

7

7

7

7

7

7

7

7

y

19

13

9

7

7

9

13

19

27

Points to be plotted :

(-4, 19) (-3, 13) (-2, 9) (-1, 7) (0, 7) (1, 9) (2, 13) (3, 19) (4, 27)

To find the x-coordinate of the vertex of the parabola, we may use the formula x = -b/2a

x = -1/2(1)  =  -1/2

By applying x = -1/2, we get the value of y.

y = (-1/2)2 + (-1/2) + 7 

y = (1/4) - (1/2) + 7

y = (1 - 2 + 28)/4  =  27/4

Vertex (1/2, 27/4)

The graph of the given parabola does not intersect the x-axis at any point. Hence it has no real roots.

(iv)  x2 − 9 = 0

Solution :

Let us give some random values of x and find the values of y.

x

-4

-3

-2

-1

0

1

2

3

4

x2

16

9

4

1

0

1

4

9

16

-9

-9

-9

-9

-9

-9

-9

-9

-9

-9

y

7

0

-5

-8

-9

-8

-5

0

7

Points to be plotted :

(-4, 7) (-3, 0) (-2, -5) (-1, -8) (0, -9) (1, -8) (2, -5) (3, 0) (4, 7)

To find the x-coordinate of the vertex of the parabola, we may use the formula x = -b/2a

x = 0/2(1)  =  0

By applying x = 0, we get the value of y.

y = 02 - 9

y = -9

Vertex (0, -9)

The graph of the given parabola intersects the x-axis at two points. Hence it has two real and unequal roots.

After having gone through the stuff given above, we hope that the students would have understood, "Finding Nature of Quadratic Equation by Graphing". 

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