You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.15

10th Maths Exercise 3.15 Solutions Question 1.
Graph the following quadratic equations and state their nature of solutions,
(i) x2 – 9x + 20 = 0
Solution:
10th Maths Exercise 3.15 Solutions Samacheer Kalvi Chapter 3 Algebra
Step 1:
Points to be plotted : (-4, 72), (-3, 56), (-2, 42), (-1, 30), (0, 20), (1, 12), (2, 6), (3, 2), (4, 0)
Step 2:
The point of intersection of the curve with x axis is (4, 0)
Step 3:
10th Maths Graph 3.15 Answers Samacheer Kalvi Chapter 3 Algebra
The roots are real & unequal
∴ Solution {4, 5}

(ii) x2 – 4x + 4 = 0
10th Maths Guide Graph Samacheer Kalvi Chapter 3 Algebra
Step 1: Points to be plotted : (-4, 36), (-3, 25), (-2, 16), (-1, 9), (0, 4), (1, 1), (2, 0), (3, 1), (4, 4)
Step 2: The point of intersection of the curve with x axis is (2, 0)
Step 3:
10th Graph Exercise 3.15 Solutions Samacheer Kalvi Chapter 3 Algebra
Since there is only one point of intersection with x axis, the quadratic equation x2 – 4x + 4 = 0 has real and equal roots.
∴ Solution{2, 2}

(iii) x2 + x + 7 = 0
Let y = x2 + x + 7
Step 1:
10th Maths Graph Answers Samacheer Kalvi Chapter 3 Algebra
Step 2:
Points to be plotted: (-4, 19), (-3, 13), (-2, 9), (-1, 7), (0, 7), (1, 9), (2, 13), (3, 19), (4, 27)
Step 3:
Draw the parabola and mark the co-ordinates of the parabola which intersect with the x-axis.
10th Class Maths Graph Pdf Samacheer Kalvi Chapter 3 Algebra
Step 4:
The roots of the equation are the points of intersection of the parabola with the x axis. Here the parabola does not intersect the x axis at any point.
So, we conclude that there is no real roots for the given quadratic equation,

(iv) x2 – 9 = 0
Let y = x2 – 9
Step 1:
10th Maths Graph Samacheer Kalvi Chapter 3 Algebra
Step 2:
The points to be plotted: (-4, 7), (-3, 0), (-2, -5), (-1, -8), (0, -9), (1,-8), (2, -5), (3, 0), (4, 7)
Step 3:
Draw the parabola and mark the co-ordinates of the parabola which intersect the x-axis.
10th Maths Exercise 3.15 Graph Samacheer Kalvi Chapter 3 Algebra
Step 4:
The roots of the equation are the co-ordinates of the intersecting points (-3, 0) and (3, 0) of the parabola with the x-axis which are -3 and 3 respectively.
Step 5:
Since there are two points of intersection with the x axis, the quadratic equation has real and unequal roots.
∴ Solution{-3, 3}

(v) x2 – 6x + 9 = 0
Let y = x2 – 6x + 9
Step 1:
10th Graph Samacheer Kalvi Chapter 3 Algebra
Step 2:
Points to be plotted: (-4, 49), (-3, 36), (-2, 25), (-1, 16), (0, 9), (1, 4), (2, 1), (3, 0), (4, 1)
Step 3:
Draw the parabola and mark the co-ordinates of the intersecting points.
Samacheer Kalvi 10th Maths Book Graph Solutions Chapter 3 Algebra
Step 4:
The point of intersection of the parabola with x axis is (3, 0)
Since there is only one point of intersection with the x-axis, the quadratic equation has real and equal roots. .
∴ Solution (3, 3)

(vi) (2x – 3)(x + 2) = 0
2x2 – 3x + 4x – 6 = 0
2x2 + 1x – 6 = 0
Let y = 2x2 + x – 6 = 0
Step 1:
Samacheer Kalvi 10th Maths Book Graph Solution Chapter 3 Algebra
Step 2:
The points to be plotted: (-4, 22), (-3, 9), (-2, 0), (-1, -5), (0, -6), (1, -3), (2, 4), (3, 15), (4, 30)
Step 3:
Draw the parabola and mark the co-ordinates of the intersecting point of the parabola with the x-axis.
10th Maths Exercise 3.15 Samacheer Kalvi Chapter 3 Algebra
Step 4:
The points of intersection of the parabola with the x-axis are (-2, 0) and (1.5, 0).
Since the parabola intersects the x-axis at two points, the, equation has real and unequal roots.
∴ Solution {-2, 1.5}

Question 2.
Draw the graph of y = x2 – 4 and hence solve x2 – x – 12 = 0
Solution:
10th Maths 3.15 Graph Samacheer Kalvi Solutions Chapter 3 Algebra
Ex 3.15 Class 10 Samacheer Kalvi Solutions Chapter 3 Algebra
10th Maths Exercise 3.15 Samacheer Kalvi Chapter 3 Algebra
Point of intersection (-3, 5), (4, 12) solution of x2 – x – 12 = 0 is -3, 4

10th Maths Graph 3.15 Answers Question 3.
Draw the graph of y = x2 + x and hence solve x2 + 1 = 0.
Solution:
Samacheer Kalvi 10th Maths Chapter 3 Algebra Ex 3.15 16
Draw the parabola by the plotting the points (-4, 12), (-3, 6), (-2, 2), (-1, 0), (0, 0), (1, 2), (2, 6), (3, 12), (4, 20), (5, 30)
10th Maths 3.15 Samacheer Kalvi Solutions Chapter 3 Algebra
To solve: x2 + 1 = 0, subtract x2 + 1 = 0 from y = x2 + x.
x2 + 1 = 0 from y = x2 + x
Samacheer Kalvi 10th Maths Exercise 3.15 Algebra
Plotting the points (-2, -3), (0, -1), (2, 1) we get a straight line. This line does not intersect the parabola. Therefore there is no real roots for the equation x2 + 1 = 0.

10th Maths Guide Graph Question 4.
Draw the graph of y = x2 + 3x + 2 and use it to solve x2 + 2x + 1 = 0.
Solution:
10th Graph Exercise 3.15 Solutions In Tamil Chapter 3 Algebra
Draw the parabola by plotting the point (-4, 6), (-3, 2), (-2, 0), (-1, 0), (0, 2), (1, 6), (2, 12), (3, 20), (4, 30).
Graph 10th Maths Samacheer Kalvi Chapter 3 Algebra Ex 3.15
To solve x2 + 2x + 1 = 0, subtract x2 + 2x + 1 = 0 from y = x2 + 3x + 2
Exercise 3.15 Class 10 Samacheer Kalvi Chapter 3 Algebra
Draw the straight line by plotting the points (-2, -1), (0, 1), (2, 3)
The straight line touches the parabola at the point (-1,0)
Therefore the x coordinate -1 is the only solution of the given equation

10th Graph Exercise 3.15 Solutions Question 5.
Draw the graph of y = x2 + 3x – 4 and hence use it to solve x2 + 3x – 4 = 0. y = x2 + 3x – 4
Solution:
10th New Syllabus Maths Graph Exercise 3.15 Samacheer Kalvi
Draw the parabola using the points (-4, 0), (-3, -4), (-2, -6), (-1, -6), (0, -4), (1, 0), (2, 6), (3, 14), (4, 24).
Samacheer Kalvi 10th Maths Chapter 3 Algebra Ex 3.15 23
To solve: x2 + 3x – 4 = 0 subtract x2 + 3x – 4 = 0 from y = x2 + 3x – 4 ,
Samacheer Kalvi 10th Maths Chapter 3 Algebra Ex 3.15 24
The points of intersection of the parabola with the x axis are the points (-4, 0) and (1, 0), whose x – co-ordinates (-4, 1) is the solution, set for the equation x2 + 3x – 4 = 0.

10th Maths Graph Answers Question 6.
Draw the graph of y = x2 – 5x – 6 and hence solve x2 – 5x – 14 = 0.
Solution:
Samacheer Kalvi 10th Maths Chapter 3 Algebra Ex 3.15 25
Draw the parabola using the points (-5, 44), (-4, 30), (-3, 18), (-2, 8), (-1, 0), (0, -6), (1, -10), (2, -12), (3, -12), (4, -10)
Samacheer Kalvi 10th Maths Chapter 3 Algebra Ex 3.15 26
To solve the equation x2 – 5x – 14 = 0, subtract x2 – 5x – 14 = 0 from y = x2 – 5x – 6.
Samacheer Kalvi 10th Maths Chapter 3 Algebra Ex 3.15 27
The co-ordinates of the points of intersection of the line and the parabola forms the solution set for the equation x2 – 5x – 14 = 0.
∴ Solution {-2, 7}

10th Class Maths Graph Pdf Question 7.
Draw the graph of y = 2x2 – 3x – 5 and hence solve 2x2 – 4x – 6 = 0. y = 2x2 – 3x – 5
Solution:
Samacheer Kalvi 10th Maths Chapter 3 Algebra Ex 3.15 28
Draw the parabola using the points (-4, 39), (-3, 22), (-2, 9), (-1, 0), (0, -5), (1, -6), (2, -3), (3, 4), (4, 15).
Samacheer Kalvi 10th Maths Chapter 3 Algebra Ex 3.15 29
To solve 2x2 – 4x – 6 = 0, subtract it from y = 2x2 – 3x – 5
Samacheer Kalvi 10th Maths Chapter 3 Algebra Ex 3.15 30
Draw a straight line using the points (-2, -1), (0, 1), (2, 3). The points of intersection of the parabola and the straight line forms the roots of the equation.
The x-coordinates of the points of intersection forms the solution set.
∴ Solution {-1, 3}

10th Maths Graph Question 8.
Draw the graph of y = (x – 1)(x + 3) and hence solve x2 – x – 6 = 0.
Solution:
y = (x – 1)(x + 3) = x2 – x + 3x – 3 = 0
y = x2 + 2x – 3
Samacheer Kalvi 10th Maths Chapter 3 Algebra Ex 3.15 31
Draw the parabola using the points (-4, 5), (-3, 0), (-2, -3), (-1,-4), (0, -3), (1, 0), (2, 5), (3, 12), (4, 21)
Samacheer Kalvi 10th Maths Chapter 3 Algebra Ex 3.15 32
To solve the equation x2 – x – 6 = 0, subtract x2 – x – 6 = 0 from y = x2 – 2x – 3.
Samacheer Kalvi 10th Maths Chapter 3 Algebra Ex 3.15 33
Plotting the points (-2, -3), (-1, 0), (0, 3), (2, 9), we get a straight line.
The points of intersection of the parabola with the straight line gives the roots of the equation. The co¬ordinates of the points of intersection forms the solution set.
∴ Solution {-2, 3}

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