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Published on: 13/05/2022
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Questions + Answers key
Take MCQ Maths Test1.
Draw the graph of y = x2 - 4x + 3 and use it to solve x2 - 6x + 9 = 0
2.
Draw the graph of y = x2 + x - 2 and hence solve x2 + x - 2 = 0
3.
Draw the graph of y = x2 + 4x + 3 and hence find the roots of x2 + x + 1 = 0
4.
Draw the graph of y = 2x2 and hence solve 2x2 - x - 6 = 0
5.
Discuss the nature of solutions of the following quadratic equations.
x2 + x - 12 = 0
1.
Step 1 : Draw the graph of y = x2 - 4x + 3 by preparing the table of values as below
| x | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
| y | 15 | 8 | 3 | 0 | -1 | 0 | 3 |
Step 2 : To solve x2 - 6x + 9 = 0, subtract x2 - 6x + 9 = 0 from y = x2 - 4x + 3
The equation y = 2x - 6 represent a straight line. Draw the graph of y = 2x - 6 forming the table of values as below.
| x | 0 | 1 | 2 | 3 | 4 | 5 |
| y | -6 | -4 | -2 | 0 | 2 | 4 |
The line y = 2x - 6 intersect y = x2 - 4x + 3 only at one point.
Step 3 : Mark the point of intersection of the curve y = x2 - 4x + 3 and y = 2x - 6 that is (3,0).
Therefore, the x coordinate 3 is the only solution for the equation x2 - 6x + 9 = 0
2.
Step 1 : Draw the graph of y = x2 + x - 2 by preparing the table of values as below
| x | -3 | -2 | -1 | 0 | 1 | 2 |
| y | 4 | 0 | -2 | -2 | 0 | 4 |
Step 2 : To solve x2 + x - 2 = 0 subtract x2 + x - 2 = 0 from y = x2 + x - 2
The equation y = 0 represents the X axis.
Step 3 : Mark the point of intersection of the curve x2 + x - 2 with the X axis. That is (–2,0) and (1,0)
Step 4 : The x coordinates of the respective points form the solution set {−2,1} for x2 + x - 2 = 0
3.
Step 1 : Draw the graph of y = x2 + 4x + 3 by preparing the table of values as below
| x | -4 | -3 | -2 | -1 | 0 | 1 | 2 |
| y | 3 | 0 | -1 | 0 | 3 | 8 | 15 |
Step 2 : To solve x2 + x + 1 = 0, subtract x2 + x + 1 = 0 from y = x2 + 4x + 3 that is,
The equation represent a straight line. Draw the graph of y = 3x + 2 forming the table of values as below
| x | -2 | -1 | 0 | 1 | 2 |
| y | -4 | -1 | 2 | 5 | 3 |
Step 3 : Observe that the graph of y = 3x + 2 does not intersect or touch the graph of the parabola y = x2 + 4x + 3.
Thus x2 + x + 1 = 0 has no real roots.
4.
Step 1: Draw the graph of y = 2x2 by preparing the table of values as below
| x | -2 | -1 | 0 | 1 | 2 |
| y | 3 | 2 | 0 | 2 | 8 |
Step 2 : To solve 2x2 - x - 6 = 0, subtract 2x2 - x - 6 = 0 from y = 2x2
The equation y = x + 6 represents a straight line. Draw the graph of y = x + 6 by forming table of values as below
| x | -2 | -1 | 0 | 1 | 2 |
| y | 4 | 5 | 6 | 7 | 8 |
Step 3 : Mark the points of intersection of the curve y = 2x2 and the line y = x + 6. That is, (–1.5, 4.5) and (2,8)
Step 4 : The x coordinates of the respective points forms the solution set {–1.5,2} for 2x2 - x - 6 = 0
5.
x2 + x - 12 = 0
Step 1 Prepare the table of values for the equation y = x2 + x - 12
| x | -5 | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
| y | 8 | 0 | -6 | -10 | -12 | -12 | -10 | -6 | 0 | 8 |
Step 2: Plot the points for the above ordered pairs (x, y) on the graph using suitable scale.
Step 3: Draw the parabola and mark the co-ordinates of the parabola which intersect the X axis.
Step 4: The roots of the equation are the x coordinates of the intersecting points (–4, 0) and (3,0)of the parabola with the X axis which are −4 and 3 respectively.
Since there are two points of intersection with the X axis, the quadratic equation x2 + x - 12 = 0 has real and unequal roots
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Tamilnadu Stateboard 10th Standard Subjects
Tamilnadu Stateboard Standards
