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.
Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Sep 07, 23 12:26 AM
Sep 06, 23 10:42 PM
Sep 06, 23 10:56 AM
