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.
Graph the following quadratic equations and state their nature of solutions.
Question 1 :
x2 - 6x + 9 = 0
Solution :
Draw the graph for the function y = x2 - 6x + 9.
Let us substitute some random values for x and find the corresponding values of y.
x -3 -2 -1 0 1 2 3 4 |
x2 9 4 1 0 1 4 9 16 |
-6x -18 -12 -6 0 -6 -12 -18 -24 |
+9 9 9 9 9 9 9 9 9 |
y 0 1 4 9 4 1 0 1 |
Points to be plotted :
(-3, 0), (-2, 1), (-1, 4), (0, 9), (1, 4), (2, 1), (3, 0), (4, 1)
Formula 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
2x2 + 4x - 3x - 6 = 0
2x2 + x - 6 = 0
Let us give some random values of x and find the values of y.
y = 2x2 + x - 6 :
x -4 -3 -2 -1 0 1 2 3 4 |
2x2 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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