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

Question 1.
Graph the following quadratic equations and state their nature of solutions.
(i) x2 – 9x + 20 = 0
(ii) x2 – 4x + 4 = 0
(iii) x2 + x + 7 = 0
(iv) x2 – 9 = 0
(v) x2 – 6x + 9 = 0
(vi) (2x – 3) (x + 2) = 0

(i) x2 – 9x + 20 = 0
Let y = x2 – 9x + 20
(i) Prepare the table of values for y = x2 – 9x + 20

(ii) Plot the points (-1, 30) (0,20) (1, 12) (2, 6) (3,2), (4, 0), (5, 0), (6,2) (omit the high value)
(iii) Join the points by a free hand smooth curve.
(iv) The roots of the equation are the X-coordinates of the intersecting points of the curve with X-axis (4, 0) and (5, 0)
There are two points of intersection with the X-axis at 4 and 5. The solution set is 4 and 5. The quadratic equation has real and unequal roots.
(v) Since there is two point of intersection with X-axis (different solution)
∴ The equation x2 – 9x + 20 = 0 has real and unequal roots.

(ii) x2 – 4x + 4 = 0
Let y = x2 – 4x + 4
(i) Prepare the table of values for y = x2 – 4x + 4

(ii) Plot the points (-3,25) (-2,16) (-1, 9) (0,4) (1,-1) (2, 0), (3,1) and (4, 4)
(iii) Join the points by a free hand smooth curve.
(iv) The roots of the equation are the X-coordinates of the intersecting points of the curve with X-axis (2, 0) which is 2.
(v) Since there is only one point of intersection with X-axis (2, 0).
∴ The solution set is 2.
The Quadratic equation x2 – Ax + 4 = 0 has real and equal roots.

(iii) x2 + x + 7 = 0
Let y = x2 + x + 7
(i) Prepare the table of values for y = x2 + x + 7

(ii) Plot the points (-4,19) (-3,13) (-2, 9) (-1, 7) (0, 7) (1, 9), (2,13) (3,19) and (4,27)
(iii) Join the points by a free hand smooth curve.
(iv) The solution of the given quadratic equation are the X-coordinates of the intersecting points of the parabola with the X-axis.
(v) The curve does not intersecting the X-axis. There is no solution set.
The equation x2 + x + 7 = 0 has no real roots.

(iv) x2 – 9 = 0
Let y = x2 – 9
(i) Prepare the table of values for y = x2 – 9

(ii) Plot the points (-4, 7) (-3, 0) (-2, -5) (-1, -8) (0, -9) (1, -8), (2, -5) (3, 0) (4, 7)
(iii) Join the points by a free hand smooth curve.
(iv) The curve intersect the X-axis at -3 and 3.
The solution is (-3, 3).
(v) Since there are two points of intersection -3 and 3 with the X-axis the quadratic equation has real and unequal roots.

(v) x2 – 6x + 9 = 0
Let y = x2 – 6x + 9
(i) Prepare a table of values for y = x2 – 6x + 9

(ii) Plot the points (-2,25) (-1,16) (0,9) (1,4) (2,1) (3,0), (4,1) and (5,4) on the graph using suitable scale (omit the points (-4, 49) and (-3, 36)
(iii) Join the points by a free hand smooth curve.
(iv) The X – coordinates of the point of intersection of the curve with X-axis are the roots of the , given equation, provided they intersect.
The solution is 3.
(v) Since there is only one point of intersection with X-axis the quadratic equation x2 – 6x + 9 = 0 has real and equal roots.

(vi) (2x – 3) (x + 2) = 0
y = (2x – 3) (x + 2)
= 2x2 + 4x – 3x – 6
= 2x2 + x – 6

(i) Prepare a table of values for y from x – 4 to 4

(ii) Plot the points (-4, 22) (-3, 9) (-2, 0) (-1, -5) (0, -6) (1, -3), (2, 4), (3, 15) and (4, 30).
(iii) Join the points by a free hand smooth curve.
(iv) The curve intersect the X – axis at (-2, 0) and (1$$\frac { 1 }{ 2 }$$, 0)
∴ The solution set is (-2,1$$\frac { 1 }{ 2 }$$)
(v) Since there are two points of intersection with X – axis, the quadratic equation has real and un – equal roots.

Question 2.
Draw the graph of y = x2 – 4 and hence solve x2 – x – 12 = 0
(i) Draw the graph of y = x2 – 4 by preparing the table of values as below.

(ii) Plot the points for the ordered pairs (-4, 12) (-3, 5) (-2, 0) (-1, -3) (0, -4) (1, -3), (2, 0), (3, 5) and (4, 12). Draw the curve with the suitable scale.
(iii) To solve x2 – x – 12 = 0 subtract x2 – x – 12 from y = x2 – 4

The equation y = x + 8 represents a straight line. Prepare a table for y = x + 8

(iv) Mark the point of intersection of the curve and the straight line is (-3, 5) and 4, 12)
∴ The solution set is (-3, 4) for x2 – x – 12 = 0.

Question 3.
Draw the graph of y = x2 + x and hence solve x2 + 1 = 0
Let y = x2 + x
(i) Draw the graph of y = x2 + x by preparing the table.

(ii) Plot the points (-4, 12), (-3, 6), (-2, 2), (-1, 0), (0, 0), (1, 2), (2, 6), (3, 12) and (4, 20).
(iii) Join the points by a free hand to get smooth curve.
(iv) To solve x2 + 1 = 0, subtract x2 + 1 = 0 from x2 + x we get.

The equation represent a straight line. Draw a line y = x – 1

Observe the graph of y = x2 + 1 does not interset the parabola y = x2 + x.
This x2 + 1 has no real roots.

Question 4.
Draw the graph of y = x2 + 3x + 2 and use it to solve x2 + 2x + 1 = 0.
(i) Draw the graph of y = x2 + 3x + 2 preparing the table of values as below.

(ii) Plot the points (-4, 6), (-3, 2), (-2, 0), (-1, 0), (0, 2), (1, 6), (2, 12), (3, 20) (4, 30).
(iii) To solve x2 + 2x + 1 = 0 subtract x2 + 2x + 1 = 0 from y = x2 + 3x + 2

(iv) Draw the graph of y = x + 1 from the table

The equation y = x + 1 represent a straight line.
This line intersect the curve at only one point (-1, 0). The solution set is (-1).

Question 5.
Draw the graph of y = x2 + 3x – 4 and hence use it to solve x2 + 3x – 4 = 0
Let y = x2 + 3x – 4
(i) Draw the graph of y = x2 + 3x – 4

(ii) Plot the points (-5, 6), (-4, 0), (-3, -4), (-2, -6), (-1, -6), (0, -4), (1, 0), (2, 6), (3, 14) on the graph using suitable scale.
(iii) Join the points by a free hand smooth curve.
The smooth curve is the graph of y = x2 – 4x + 4
(iv) To solve x + 3x – 4 = 0, subtract x2 + 3x – 4 = 0 from y = x2 + 3x – 4.
y = 0
∴ The point of intersection with the x – axis is the solution set.
The solution set is -4 and 1.

Question 6.
Draw the graph of y = x2 – 5x – 6 and hence solve x2, – 5x – 14 = 0
Let y = x2 – 5x – 6
(i) Draw the graph of y = x2 – 5x – 6 by preparing the table of values as below.

(ii) Plot the points (-3, 18), (-2, 8), (-1, 0), (0, -6), (1, -10), (2, -12), (3, -12), (4, -10), (5,-6), (6, 0) and (7, 8).
(iii) Join the points by a free hand to get smooth curve.
(iv) To solve x2 – 5x – 14 = 0, subtract x2 – 5x – 14 = 0 from y = x2 – 5x – 6.

The equation y = 8 represent a straight line draw a straight line through y = 8 intersect the curve at two places. From the two points draw perpendicular line to the X – axis it will intersect at -2 and 7.
The solution is -2 and 7

Question 7.
Draw the graph of y = 2x2 – 3x – 5 and hence solve 2x2 – 4x – 6 = 0
(i) Draw the graph of y = 2x2 – 3x – 5 by preparing the table of values given below.

(ii) Plot the points (-3, 22), (-2, 9), (-1, 0), (0, -5), (1,-6), (2, -3), (3, 4), (4, 15) on the graph sheet using suitable scale.
(iii) To solve 2x2 – 4x – 6 = 0 subtract 2x2 – 4x – 6 = 0 from y = 2x2 – 3x – 5

(iv) y = x + 1 represent a straight line.

The straight line intersect the curve at (-1, 0) and (3, 4). From the two point draw perpendicular lines to the X – axis it will intersect at -1 and 3.
The solution set is (-1, 3)

Question 8.
Draw the graph of y = (x – 1) (x + 3) and hence solve x2 – x – 6 = 0