OP Malhotra Class 11 Maths Solutions Chapter 10 Quadratic Equations Exercise 10 (E)

 

Question 1. y = x² – 5x + 6 ; 0 ≤ x ≤ 5.
Answer: To draw the graph for the quadratic function \( y = x^2 - 5x + 6 \) within the range \( 0 \le x \le 5 \), we first calculate a table of values.
The table of values is given as under:

From the graph, we can clearly see that the curve crosses the x-axis at two points: \( x = 2 \) and \( x = 3 \). These are the solutions to the equation. A parabola will often intersect the x-axis at two points.
To verify this algebraically:
Given equation: \( y = x^2 - 5x + 6 \)
We can factorize the quadratic expression:
\( y = x^2 - 2x - 3x + 6 \)
\( = x(x - 2) - 3(x - 2) \)
\( \implies y = (x - 2)(x - 3) \)
For \( y = 0 \), we have \( (x - 2)(x - 3) = 0 \)
This means either \( x - 2 = 0 \) or \( x - 3 = 0 \).
So, \( x = 2 \) or \( x = 3 \).
These algebraic solutions match the points where the graph intersects the x-axis.
In simple words: First, we make a table of points by putting different 'x' values into the equation to find 'y' values. Then, we plot these points on a graph and draw a curve through them. Where this curve touches the 'x' line, those are our answers for 'x'. We can also check this by breaking down the equation using multiplication.

🎯 Exam Tip: When plotting quadratic graphs, make sure to include points that show the turning point (vertex) of the parabola, as this is crucial for the curve's shape. Also, clearly mark the x-intercepts if they exist.

 

Question 2. y = − x² + 2x + 3 ; − 3 ≤ x ≤ 5.
Answer: To graph the function \( y = -x^2 + 2x + 3 \) for \( -3 \le x \le 5 \), we first calculate the corresponding y-values for various x-values.
The table of values is given as under:

Looking at the graph, the curve crosses the x-axis at \( x = -1 \) and \( x = 3 \). These are the real solutions for the quadratic equation. A parabola that opens downwards (because of the negative \( x^2 \) term) typically shows this behavior.
To verify the solutions using algebra:
Given equation: \( y = -x^2 + 2x + 3 \)
Factorizing the expression by taking out a negative sign:
\( y = -(x^2 - 2x - 3) \)
\( = -(x^2 - 3x + x - 3) \)
\( = -[x(x - 3) + 1(x - 3)] \)
\( \implies y = -(x - 3)(x + 1) \)
For \( y = 0 \), we have \( -(x - 3)(x + 1) = 0 \)
This means either \( x - 3 = 0 \) or \( x + 1 = 0 \).
So, \( x = 3 \) or \( x = -1 \).
These algebraic solutions perfectly match the points where the graph intersects the x-axis.
In simple words: We find points by putting different 'x' numbers into the equation to get 'y' numbers. We draw these points and the curve on a grid. The 'x' values where the curve touches the horizontal 'x' line are the answers. We can double-check this by taking a minus sign out of the equation and then splitting it into two simpler parts that multiply together.

🎯 Exam Tip: When the leading coefficient of a quadratic equation is negative (like \( -x^2 \)), the parabola opens downwards, indicating a maximum point instead of a minimum. Ensure your table of values includes the vertex to accurately plot the curve's turning point.

 

Question 3. y = x² – 4x + 4 ;-1 ≤ x ≤ 5.
Answer: To construct the graph for \( y = x^2 - 4x + 4 \) within the specified range of \( -1 \le x \le 5 \), we must first determine a series of coordinate pairs.
The table of values is given as under:

From the graph, it is clear that the curve touches the x-axis at only one point, which is \( x = 2 \). This means the quadratic equation has exactly one real solution, or two identical real solutions. The parabola has its vertex on the x-axis.
Algebraic verification:
Given equation: \( y = x^2 - 4x + 4 \)
This is a perfect square trinomial:
\( y = (x - 2)^2 \)
For \( y = 0 \), we have \( (x - 2)^2 = 0 \)
Taking the square root of both sides gives \( x - 2 = 0 \).
So, \( x = 2 \).
This algebraic solution confirms that the graph touches the x-axis at only one point, \( x = 2 \).
In simple words: We list out different 'x' values and calculate their 'y' values to make a table. Then, we draw these points on a graph. The curve we draw touches the 'x' line at only one place. This means 'x' has just one answer. We can also see this because the equation can be written as 'x minus 2' multiplied by itself.

🎯 Exam Tip: When a quadratic graph touches the x-axis at exactly one point, it indicates that the quadratic equation has repeated roots (or a single real root). This happens when the discriminant (\( b^2 - 4ac \)) of the quadratic formula is zero.

 

Question 4. Solve graphically and compare your answer with algebraic solution either by factorization or formula method :
Answer:
(i) \( y = x^2 - 5x + 6 \)
To solve this equation graphically, we first create a table of values for \( y = x^2 - 5x + 6 \).
The table of values is given as under:

The graph clearly shows that the curve intersects the x-axis at \( x = 2 \) and \( x = 3 \). These are the graphical solutions.
To verify algebraically:
Given equation: \( x^2 - 5x + 6 = 0 \)
By factorization:
\( x^2 - 2x - 3x + 6 = 0 \)
\( x(x - 2) - 3(x - 2) = 0 \)
\( \implies (x - 2)(x - 3) = 0 \)
So, \( x = 2 \) or \( x = 3 \).
The graphical and algebraic solutions match, which is a good way to double-check your work. Drawing a graph helps visualize how a quadratic equation behaves.

(ii) \( y = -x^2 + 2x + 3 \)
To solve this equation graphically, we use a table of values for \( y = -x^2 + 2x + 3 \).
The table of values is given as under:

From the graph, we can see that the curve intersects the x-axis at \( x = -1 \) and \( x = 3 \). These are the graphical solutions.
To verify algebraically:
Given equation: \( -x^2 + 2x + 3 = 0 \)
Multiply by -1 to make the \( x^2 \) term positive:
\( x^2 - 2x - 3 = 0 \)
By factorization:
\( x^2 - 3x + x - 3 = 0 \)
\( x(x - 3) + 1(x - 3) = 0 \)
\( \implies (x - 3)(x + 1) = 0 \)
So, \( x = 3 \) or \( x = -1 \).
Both methods yield the same solutions. Graphing the function helps to visualize the roots of the equation clearly.

(iii) \( y = x^2 - 4x + 4 \)
To solve this graphically, we generate a table of values for \( y = x^2 - 4x + 4 \).
The table of values is given as under:

The graph clearly shows that the curve intersects the x-axis at only one point, \( x = 2 \). This means there is one unique real solution.
To verify algebraically:
Given equation: \( x^2 - 4x + 4 = 0 \)
This is a perfect square:
\( (x - 2)^2 = 0 \)
Taking the square root of both sides gives \( x - 2 = 0 \).
So, \( x = 2 \).
Both graphical and algebraic methods agree on the solution. When a quadratic expression is a perfect square, its graph will just touch the x-axis at one point, which is its vertex.

(iv) \( y = x^2 - x - 6 \)
To solve this graphically, we generate a table of values for \( y = x^2 - x - 6 \).
The table of values is given as under:

The graph clearly meets the x-axis at \( (-2, 0) \) and \( (3, 0) \). So, the real solutions are \( x = -2 \) and \( x = 3 \). Graphing helps us visualize the points where the function equals zero.
To verify algebraically:
Given equation: \( x^2 - x - 6 = 0 \)
By factorization:
\( x^2 - 3x + 2x - 6 = 0 \)
\( x(x - 3) + 2(x - 3) = 0 \)
\( \implies (x + 2)(x - 3) = 0 \)
So, \( x = -2 \) or \( x = 3 \).
The algebraic solutions align with the graphical findings. This confirms the accuracy of both methods.

(v) \( y = x^2 - 6x + 9 \)
To solve this graphically, we prepare a table of values for \( y = x^2 - 6x + 9 \).
The table of values is given as under:

The graph clearly shows that the curve touches the x-axis at \( (3, 0) \). This indicates that \( x = 3 \) is the repeated solution for the equation. A parabola touching the x-axis means its vertex lies on the x-axis.
To verify algebraically:
Given equation: \( x^2 - 6x + 9 = 0 \)
This is a perfect square trinomial:
\( (x - 3)^2 = 0 \)
Taking the square root of both sides:
\( x - 3 = 0 \)
So, \( x = 3 \).
Both methods confirm that \( x = 3 \) is the solution. This consistency provides confidence in the answer.

(vi) \( y = -x^2 + x + 12 \)
To solve this graphically, we construct a table of values for \( y = -x^2 + x + 12 \).
The table of values is given as under:

The graph shows that the curve intersects the x-axis at \( x = -3 \) and \( x = 4 \). These are the real solutions for the given quadratic equation. It's a downward-opening parabola, as indicated by the negative \( x^2 \) term.
To verify algebraically:
Given equation: \( -x^2 + x + 12 = 0 \)
Multiply by -1:
\( x^2 - x - 12 = 0 \)
By factorization:
\( x^2 - 4x + 3x - 12 = 0 \)
\( x(x - 4) + 3(x - 4) = 0 \)
\( \implies (x + 3)(x - 4) = 0 \)
So, \( x = -3 \) or \( x = 4 \).
The graphical and algebraic solutions match, reinforcing the correctness of both approaches.

(vii) \( y = x^2 - 4x + 5 = 0 \)
To solve this graphically, we generate a table of values for \( y = x^2 - 4x + 5 \).
The table of values is given as under:

The graph clearly shows that the curve does not intersect or touch the x-axis at any point. This implies that the quadratic equation \( x^2 - 4x + 5 = 0 \) has no real solutions. Instead, it has complex or imaginary solutions.
To verify algebraically:
Given equation: \( x^2 - 4x + 5 = 0 \)
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -4, c = 5 \).
\( x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(5)}}{2(1)} \)
\( x = \frac{4 \pm \sqrt{16 - 20}}{2} \)
\( x = \frac{4 \pm \sqrt{-4}}{2} \)
\( x = \frac{4 \pm 2i}{2} \)
\( x = 2 \pm i \)
Since the solutions involve the imaginary unit \( i \), they are complex numbers. This confirms that there are no real solutions, which is consistent with the graph not intersecting the x-axis.

(viii) \( y = x^2 + 2x + 2 = 0 \)
To solve this graphically, we prepare a table of values for \( y = x^2 + 2x + 2 \).
The table of values is given as under:

The graph clearly shows that the curve does not intersect the x-axis. Therefore, the given quadratic equation \( x^2 + 2x + 2 = 0 \) has no real solutions. This is typical for parabolas that are entirely above or below the x-axis.
To verify algebraically:
Given equation: \( x^2 + 2x + 2 = 0 \)
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = 2, c = 2 \).
\( x = \frac{-2 \pm \sqrt{2^2 - 4(1)(2)}}{2(1)} \)
\( x = \frac{-2 \pm \sqrt{4 - 8}}{2} \)
\( x = \frac{-2 \pm \sqrt{-4}}{2} \)
\( x = \frac{-2 \pm 2i}{2} \)
\( x = -1 \pm i \)
Since the solutions involve the imaginary unit \( i \), they are complex numbers. This algebraically confirms that there are no real solutions, matching the graphical observation.
In simple words: For each part, we create a table of 'x' and 'y' values, then draw the curve on a graph. The points where the curve crosses the 'x' line are the solutions. If it doesn't cross, there are no real answers. We then use methods like splitting the equation into factors or the special formula to check if our answers are correct. Always compare the graph's crossing points with the numbers from algebra.

🎯 Exam Tip: When a quadratic equation has no real solutions, its graph (a parabola) will not intersect or touch the x-axis. Using the discriminant (\( b^2 - 4ac \)) from the quadratic formula can quickly tell you if real solutions exist: if it's negative, there are no real solutions.