In this section, you will learn, how to examine the nature of roots of a quadratic equation using its 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.

Question 1 :

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

x2 – 6x + 9 = 0

Solution :

Draw the graph for the function y = x2 – 6x + 9

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 -6x -24 -18 -12 -6 0 -6 -12 -18 -24 +9 9 9 9 9 9 9 9 9 9 y 1 0 1 4 9 4 1 0 1

Points to be plotted :

(-4, 1) (-3, 0) (-2, 1) (-1, 4) (0, 9) (1, 4) (2, 1) (3, 0) (4, 1)

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

x = -(-6)/2(1) = 6/2 = 3

By applying x = 3, we get the value of y.

y = 32 – 6(3) + 9

y = 9 – 18 + 9

y = 0

Vertex (3, 0)

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

Question 2 :

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

(2x – 3)(x + 2) = 0

Solution :

(2x – 3)(x + 2) = 0

2×2 + 4x – 3x – 6 = 0

2×2 + x – 6 = 0

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

y = 2×2 + x – 6

 x -4 -3 -2 -1 0 1 2 3 4 2×2 32 18 8 2 0 1 8 18 32 x -4 -3 -2 -1 0 1 2 3 4 -6 -6 -6 -6 -6 -6 -6 -6 -6 -6 y 22 9 0 -5 -6 -4 4 15 30

Points to be plotted :

(-4, 22) (-3, 9) (-2, 0) (-1, -5) (0, -6) (1, -4) (2, 4) (3, 15) (4, 30)

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

x = -1/2(2) = 1/4

By applying x = 1/4, we get the value of y.

y = 2(1/4)2 + (1/4) – 6

y = 2(1/16) + (1/4) – 6

y = -45/8

Vertex (1/4, -45/8)

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

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