# HOW TO FIND NATURE OF SOLUTION OF QUADRATIC EQUATION WITH GRAPH

How to Find Nature of Solution of Quadratic Equation with Graph ?

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.

## How to Find Nature of Solution of Quadratic Equation with Graph - Questions

Question 1 :

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

(i)  x2 − 9x + 20 = 0

Solution :

Draw the graph for the function y = x2 − 9x + 20

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

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

x = -(-9)/2(1)  =  9/2

By applying x = 9/2, we get the value of y.

y = (9/2)2 - 9(9/2) + 20

y = (81/4) - (81/2) + 20

y = (81 - 162 + 80)/4  =  -1/4

Vertex (9/2, - 1/4)

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

(ii)  x2 − 4x + 4 = 0

Solution :

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

 x-4-3-2-101234 x216941014916 -4x1612840-4-8-12-16 +4444444444 y362516941014

Points to be plotted :

(-4, 36) (-3, 25) (-2, 16) (-1, 9) (0, 4) (1, 1) (2, 0) (3, 1) (4, 4)

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

x = -(-4)/2(1)  =  4/2  =  2

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

y = 22 - 4(2) + 4

y = 4 - 8 + 4

y = 0

Vertex (2, 0)

The graph of the given parabola intersects the x-axis at only one point. it has real and equal roots.

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