Samacheer Kalvi Class 10 Maths Solutions Chapter 3 Algebra Exercise 3.15

 

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

(i) To graph \( y = x^2 - 9x + 20 \):

Plot the points from the table and join them with a smooth curve. The points where the curve crosses the X-axis are (4, 0) and (5, 0). These are the roots of the equation. Because the curve crosses the X-axis at two distinct points, the quadratic equation \( x^2 - 9x + 20 = 0 \) has real and unequal roots. The graph visually shows the solution.
In simple words: First, create a table of x and y values for the equation. Then, draw these points on a graph and connect them to form a smooth curve. Where this curve touches or crosses the straight X-line, those are the answers for x. Since it crosses twice, there are two different real answers.

🎯 Exam Tip: When plotting points for a quadratic equation, always look for the lowest or highest point (vertex) of the parabola, as this helps to understand its shape.

(ii) To graph \( y = x^2 - 4x + 4 \):

Plot the points from the table. The curve intersects the X-axis at only one point, (2, 0). This means the quadratic equation \( x^2 - 4x + 4 = 0 \) has real and equal roots. When a parabola touches the x-axis at exactly one point, it indicates a repeated root.
In simple words: Draw a table for x and y values. Plot these points and draw a smooth curve. If the curve only touches the X-line at one place, then the equation has one answer that counts as two equal answers.

🎯 Exam Tip: When the parabola touches the X-axis at only one point, it indicates that the quadratic equation has two identical real roots, also known as a repeated root.

(iii) To graph \( y = x^2 + x + 7 \):

Plot the points from the table. The curve does not intersect the X-axis at any point. This means the quadratic equation \( x^2 + x + 7 = 0 \) has no real roots. When a parabola never touches or crosses the x-axis, its solutions are complex numbers.
In simple words: Make a table for x and y values, then plot them and draw the curve. If the curve never touches or crosses the X-line, it means there are no real answers for x.

🎯 Exam Tip: When the discriminant (b² - 4ac) of a quadratic equation is negative, the parabola will never cross the X-axis, indicating no real solutions.

(iv) To graph \( y = x^2 - 9 \):

Plot the points from the table. The curve intersects the X-axis at two points: (-3, 0) and (3, 0). This means the quadratic equation \( x^2 - 9 = 0 \) has real and unequal roots. It's a simple parabola that is shifted downwards, resulting in two distinct x-intercepts.
In simple words: Create a table of values and plot them to draw the curve. The curve crosses the X-line at -3 and 3, which are the two different answers for x.

🎯 Exam Tip: Remember that equations of the form \( x^2 - c = 0 \) will always have two real roots, \( \pm \sqrt{c} \), which means the parabola will cross the X-axis symmetrically around the Y-axis.

(v) To graph \( y = x^2 - 6x + 9 \):

Plot the points from the table. The curve touches the X-axis at only one point, (3, 0). This means the quadratic equation \( x^2 - 6x + 9 = 0 \) has real and equal roots. This equation is a perfect square trinomial, which always results in a single, repeated root.
In simple words: Draw a table for x and y values. Plot these points and draw a smooth curve. The curve only touches the X-line at 3, so the equation has one answer that is counted twice.

🎯 Exam Tip: Recognizing perfect square trinomials like \( (x-3)^2 \) immediately tells you there will be one repeated root, making graph interpretation easier.

(vi) To graph \( y = (2x – 3) (x + 2) = 2x^2 + x - 6 \):

Plot the points from the table. The curve intersects the X-axis at two points: (-2, 0) and approximately (1.5, 0). This means the quadratic equation \( 2x^2 + x - 6 = 0 \) has real and unequal roots. Expanding the given factors first helps in creating the table of values accurately.
In simple words: First, multiply the brackets to get the equation in \( ax^2 + bx + c \) form. Then, make a table of values and draw the curve. The curve crosses the X-line at two different points, meaning there are two distinct real answers.

🎯 Exam Tip: When given an equation in factored form, \( (ax+b)(cx+d) = 0 \), setting each factor to zero directly provides the x-intercepts, which are the roots.

 

Question 2. Draw the graph of \( y = x^2 - 4 \) and hence solve \( x^2 – x – 12 = 0 \)
Answer:

Step 1: Graph \( y = x^2 - 4 \)

Step 2: Solve \( x^2 – x – 12 = 0 \) graphically.
We compare the equation \( x^2 – x – 12 = 0 \) with \( y = x^2 - 4 \):
\( y = x^2 - 4 \)
\( 0 = x^2 - x - 12 \)
Subtracting the two equations gives:
\( y - 0 = (x^2 - 4) - (x^2 - x - 12) \)
\( \implies y = x^2 - 4 - x^2 + x + 12 \)
\( \implies y = x + 8 \)
This new equation, \( y = x + 8 \), represents a straight line. Now, we create a table of values for this line:

Step 3: Plot both the parabola and the straight line.
The points of intersection between the parabola \( y = x^2 - 4 \) and the line \( y = x + 8 \) are (-3, 5) and (4, 12). The x-coordinates of these intersection points are the solutions to the equation \( x^2 – x – 12 = 0 \). Therefore, the solution set is (-3, 4). The graphical method is a strong visual tool for understanding quadratic roots.
In simple words: First, plot the curve for \( y = x^2 - 4 \). Then, find a straight line equation by subtracting the solving equation from the graphing equation. Plot this straight line on the same graph. The x-values where the line and curve meet are the answers.

 

Question 3. Draw the graph of \( y = x^2 + x \) and hence solve \( x^2 + 1 = 0 \)
Answer:
(i) First, we need to draw the graph for the equation \( y = x^2 + x \). To do this, we prepare a table of values:

(ii) Now, plot these points on a graph sheet: (-4, 12), (-3, 6), (-2, 2), (-1, 0), (0, 0), (1, 2), (2, 6), (3, 12), and (4, 20).
(iii) Connect these points with a smooth, freehand curve to form the parabola. This curve shows us all possible values for \( y = x^2 + x \).
(iv) To solve \( x^2 + 1 = 0 \), we compare it with our graph equation \( y = x^2 + x \). We do this by subtracting the equation to be solved from the graph equation:
\( y = x^2 + x \)
\( 0 = x^2 + 1 \)
Subtracting these gives:
\( y - 0 = (x^2 + x) - (x^2 + 1) \)
\( y = x - 1 \)
This new equation \( y = x - 1 \) represents a straight line. We need to draw this line on the same graph. First, we create a table of values for \( y = x - 1 \):

(v) Plot these points and draw the straight line. When we look at the graph, we will see that the line \( y = x - 1 \) does not touch or cross the parabola \( y = x^2 + x \) at any point.
Therefore, the equation \( x^2 + 1 = 0 \) has no real roots. This means there are no real numbers for \( x \) that would make the equation true.
In simple words: We draw the curve for the first equation. Then we find a straight line from the second equation. If this line does not cross the curve, it means there are no real solutions for the second equation.

🎯 Exam Tip: When using graphs to solve equations, always draw both the parabola and the derived straight line on the same graph sheet to clearly identify intersection points or the lack thereof.

 

Question 4. Draw the graph of \( y = x^2 + 3x + 2 \) and use it to solve \( x^2 + 2x + 1 = 0 \).
Answer:
(i) First, we need to draw the graph for the equation \( y = x^2 + 3x + 2 \). To do this, we prepare a table of values:

(ii) Plot these points on a graph sheet: (-4, 6), (-3, 2), (-2, 0), (-1, 0), (0, 2), (1, 6), (2, 12), (3, 20), (4, 30).
(iii) Connect these points with a smooth, freehand curve to form the parabola. This curve represents \( y = x^2 + 3x + 2 \).
(iv) To solve \( x^2 + 2x + 1 = 0 \), we subtract this equation from the graph equation \( y = x^2 + 3x + 2 \):
\( y = x^2 + 3x + 2 \)
\( 0 = x^2 + 2x + 1 \)
Subtracting these gives:
\( y - 0 = (x^2 + 3x + 2) - (x^2 + 2x + 1) \)
\( y = x + 1 \)
This new equation \( y = x + 1 \) represents a straight line. We will draw this line on the same graph. First, we create a table of values for \( y = x + 1 \):

(v) Plot these points and draw the straight line. This line intersects the curve at only one point, which is (-1, 0).
Therefore, the solution set for \( x^2 + 2x + 1 = 0 \) is \( \{-1\} \). This means \( x = -1 \) is the only root, and it is a repeated root.
In simple words: We draw the curve for the given equation. Then, we find a straight line from the equation we want to solve. The point where this line touches the curve gives us the answer.

🎯 Exam Tip: When the straight line is tangent to (touches at one point) the parabola, it indicates that the quadratic equation has real and equal roots (a repeated root).

 

Question 5. Draw the graph of \( y = x^2 + 3x - 4 \) and hence use it to solve \( x^2 + 3x - 4 = 0 \).
Answer:
(i) First, we need to draw the graph for the equation \( y = x^2 + 3x - 4 \). To do this, we prepare a table of values:

(ii) Plot these points on a graph sheet: (-5, 6), (-4, 0), (-3, -4), (-2, -6), (-1, -6), (0, -4), (1, 0), (2, 6), (3, 14).
(iii) Connect these points with a smooth, freehand curve to form the parabola representing \( y = x^2 + 3x - 4 \).
(iv) To solve \( x^2 + 3x - 4 = 0 \), we subtract this equation from the graph equation \( y = x^2 + 3x - 4 \):
\( y = x^2 + 3x - 4 \)
\( 0 = x^2 + 3x - 4 \)
Subtracting these equations gives:
\( y - 0 = (x^2 + 3x - 4) - (x^2 + 3x - 4) \)
\( y = 0 \)
This equation \( y = 0 \) represents the X-axis itself. So, we are looking for the points where the parabola intersects the X-axis.
The points of intersection are (-4, 0) and (1, 0).
Therefore, the solution set for \( x^2 + 3x - 4 = 0 \) is \( \{-4, 1\} \). The roots are \( x = -4 \) and \( x = 1 \).
In simple words: When the equation we want to solve is the same as the equation for which we draw the graph, the solutions are simply where the curve crosses the main horizontal line (the X-axis).

🎯 Exam Tip: When solving a quadratic equation by graphing its own function (\( y = ax^2 + bx + c \) and solving \( ax^2 + bx + c = 0 \)), the roots are precisely the x-intercepts of the parabola.

 

Question 6. Draw the graph of \( y = x^2 - 5x - 6 \) and hence solve \( x^2 - 5x - 14 = 0 \).
Answer:
(i) First, we need to draw the graph for the equation \( y = x^2 - 5x - 6 \). To do this, we prepare a table of values:

(ii) Plot these points on a graph sheet: (-3, 18), (-2, 8), (-1, 0), (0, -6), (1, -10), (2, -12), (3, -12), (4, -10), (5, -6), (6, 0), and (7, 8).
(iii) Connect these points with a smooth, freehand curve to form the parabola representing \( y = x^2 - 5x - 6 \).
(iv) To solve \( x^2 - 5x - 14 = 0 \), we subtract this equation from the graph equation \( y = x^2 - 5x - 6 \):
\( y = x^2 - 5x - 6 \)
\( 0 = x^2 - 5x - 14 \)
Subtracting these equations gives:
\( y - 0 = (x^2 - 5x - 6) - (x^2 - 5x - 14) \)
\( y = -6 - (-14) \)
\( y = -6 + 14 \)
\( y = 8 \)
This new equation \( y = 8 \) represents a straight horizontal line. We will draw this line on the same graph.
(v) Draw the horizontal line \( y = 8 \). Then, find the points where this line intersects the parabola. From these intersection points, draw perpendicular lines to the X-axis. These lines will intersect the X-axis at \( x = -2 \) and \( x = 7 \).
Therefore, the solution set for \( x^2 - 5x - 14 = 0 \) is \( \{-2, 7\} \). The roots are \( x = -2 \) and \( x = 7 \).
In simple words: We draw the curve for the first equation. Then, by subtracting equations, we find a simple horizontal line. Where this line crosses the curve, that is our answer.

🎯 Exam Tip: Remember that a horizontal line \( y = k \) (where k is a constant) will be parallel to the X-axis. Its intersections with the parabola give the roots of the modified equation.

 

Question 7. Draw the graph of \( y = 2x^2 - 3x - 5 \) and hence solve \( 2x^2 - 4x - 6 = 0 \).
Answer:
(i) First, we need to draw the graph for the equation \( y = 2x^2 - 3x - 5 \). To do this, we prepare a table of values:

(ii) Plot these points on a graph sheet: (-3, 22), (-2, 9), (-1, 0), (0, -5), (1, -6), (2, -3), (3, 4), (4, 15).
(iii) Connect these points with a smooth, freehand curve to form the parabola representing \( y = 2x^2 - 3x - 5 \).
(iv) To solve \( 2x^2 - 4x - 6 = 0 \), we subtract this equation from the graph equation \( y = 2x^2 - 3x - 5 \):
\( y = 2x^2 - 3x - 5 \)
\( 0 = 2x^2 - 4x - 6 \)
Subtracting these equations gives:
\( y - 0 = (2x^2 - 3x - 5) - (2x^2 - 4x - 6) \)
\( y = (-3x - (-4x)) + (-5 - (-6)) \)
\( y = x + 1 \)
This new equation \( y = x + 1 \) represents a straight line. We will draw this line on the same graph. First, we create a table of values for \( y = x + 1 \):

(v) Plot these points and draw the straight line. The straight line intersects the parabola at two points: (-1, 0) and (3, 4). Drawing perpendicular lines from these points to the X-axis, we find the roots are \( x = -1 \) and \( x = 3 \).
Therefore, the solution set for \( 2x^2 - 4x - 6 = 0 \) is \( \{-1, 3\} \).
In simple words: First, we draw the graph for the more complex equation. Then, we use subtraction to find a simpler straight line equation. The points where this line crosses the curve tell us the answers to the second equation.

🎯 Exam Tip: When the coefficients in the parabola's equation are larger, make sure to set an appropriate scale on your graph to fit all points accurately.

 

Question 8. Draw the graph of \( y = (x - 1)(x + 3) \) and hence solve \( x^2 - x - 6 = 0 \).
Answer:
(i) First, we need to simplify the given equation for the graph: \( y = (x - 1)(x + 3) = x^2 + 3x - x - 3 = x^2 + 2x - 3 \). So, we will draw the graph for \( y = x^2 + 2x - 3 \). To do this, we prepare a table of values:

(ii) Plot these points on a graph sheet: (-4, 5), (-3, 0), (-2, -3), (-1, -4), (0, -3), (1, 0), (2, 5), (3, 12), and (4, 21).
(iii) Connect these points with a smooth, freehand curve to form the parabola representing \( y = x^2 + 2x - 3 \).
(iv) To solve \( x^2 - x - 6 = 0 \), we subtract this equation from the graph equation \( y = x^2 + 2x - 3 \):
\( y = x^2 + 2x - 3 \)
\( 0 = x^2 - x - 6 \)
Subtracting these equations gives:
\( y - 0 = (x^2 + 2x - 3) - (x^2 - x - 6) \)
\( y = (2x - (-x)) + (-3 - (-6)) \)
\( y = 3x + 3 \)
This new equation \( y = 3x + 3 \) represents a straight line. We will draw this line on the same graph. First, we create a table of values for \( y = 3x + 3 \):

(v) Plot these points and draw the straight line. This straight line cuts the parabola at two points: (-2, -3) and (3, 12). Drawing perpendicular lines from these points to the X-axis, we find the roots are \( x = -2 \) and \( x = 3 \).
Therefore, the solution set for \( x^2 - x - 6 = 0 \) is \( \{-2, 3\} \).
In simple words: We first simplify the equation for the graph and then plot its curve. Next, we use the equation we want to solve to find a straight line. The points where this line crosses the curve will give us the answers to the problem.

🎯 Exam Tip: Always expand and simplify factored quadratic equations like \( y = (x-1)(x+3) \) before creating the table of values to avoid errors in calculation.