OP Malhotra Class 10 Maths Solutions Chapter 5 Quadratic Equations Exercise 5 (D)

Question 1. Use the discriminant to determine the nature of the roots of each of the following quadratic equations :
(i) \( x^2 - 4x + 3 = 0 \)
(ii) \( x^2 - 4x + 5 = 0 \)
(iii) \( x^2 - 4x + 4 = 0 \)
Answer:
(i) For the equation \( x^2 - 4x + 3 = 0 \):
First, we identify the coefficients: \( a = 1 \), \( b = -4 \), and \( c = 3 \).
Next, we calculate the discriminant \( D = b^2 - 4ac \).
\( D = (-4)^2 - 4(1)(3) \)
\( D = 16 - 12 \)
\( D = 4 \)
Since \( D = 4 \) is greater than 0 (\( D > 0 \)), the roots are real and distinct. Also, because 4 is a perfect square, the roots are rational, meaning they can be expressed as simple fractions.
(ii) For the equation \( x^2 - 4x + 5 = 0 \):
The coefficients are: \( a = 1 \), \( b = -4 \), and \( c = 5 \).
Now, we calculate the discriminant \( D = b^2 - 4ac \).
\( D = (-4)^2 - 4(1)(5) \)
\( D = 16 - 20 \)
\( D = -4 \)
Since \( D = -4 \) is less than 0 (\( D < 0 \)), the roots are not real; they are imaginary. Imaginary roots involve the square root of a negative number.
(iii) For the equation \( x^2 - 4x + 4 = 0 \):
The coefficients are: \( a = 1 \), \( b = -4 \), and \( c = 4 \).
Let's calculate the discriminant \( D = b^2 - 4ac \).
\( D = (-4)^2 - 4(1)(4) \)
\( D = 16 - 16 \)
\( D = 0 \)
Since \( D = 0 \), the roots are real and equal. This also means the quadratic equation is a perfect square.
(iv) For the equation \( x^2 - x = 7 \):
First, we rewrite the equation in the standard form \( ax^2 + bx + c = 0 \):
\( x^2 - x - 7 = 0 \)
The coefficients are: \( a = 1 \), \( b = -1 \), and \( c = -7 \).
Now, we calculate the discriminant \( D = b^2 - 4ac \).
\( D = (-1)^2 - 4(1)(-7) \)
\( D = 1 + 28 \)
\( D = 29 \)
Since \( D = 29 \) is greater than 0 (\( D > 0 \)), the roots are real and distinct. However, because 29 is not a perfect square, the roots are irrational and cannot be written as a simple fraction.
In simple words: To find out about the roots of a quadratic equation, we use a special number called the discriminant. If this number is positive and a perfect square, the roots are real, different, and can be written as fractions. If it's positive but not a perfect square, the roots are real, different, but not simple fractions. If it's zero, the roots are real and exactly the same. If it's negative, there are no real roots.

🎯 Exam Tip: Remember to always write the quadratic equation in the standard form \( ax^2 + bx + c = 0 \) before identifying a, b, and c to calculate the discriminant correctly.

Question 2. Without finding the roots, comment on the nature of the roots of the following quadratic equations.
(i) \( 3x^2 - 6x + 5 = 0 \)
(ii) \( 5y^2 + 12y - 9 = 0 \)
(iii) \( x^2 - 5x - 7 = 0 \)
(iv) \( a^2x^2 + abx = b^2, a \neq 0 \)
(v) \( 9a^2b^2x^2 - 48abcdx + 64c^2d^2 = 0, a \neq 0, b \neq0 \)
(vi) \( 4x^2 = 1 \)
(vii) \( 64x^2 - 112x + 49 = 0 \)
(viii) \( 8x^2 + 5x + 1 = 0 \)
Answer:
(i) For the equation \( 3x^2 - 6x + 5 = 0 \):
We have \( a = 3 \), \( b = -6 \), and \( c = 5 \).
The discriminant is \( D = b^2 - 4ac \).
\( D = (-6)^2 - 4(3)(5) \)
\( D = 36 - 60 \)
\( D = -24 \)
Since \( D < 0 \), the roots are not real; they are imaginary. These roots cannot be plotted on a regular number line.
(ii) For the equation \( 5y^2 + 12y - 9 = 0 \):
We have \( a = 5 \), \( b = 12 \), and \( c = -9 \).
The discriminant is \( D = b^2 - 4ac \).
\( D = (12)^2 - 4(5)(-9) \)
\( D = 144 + 180 \)
\( D = 324 \)
Since \( D > 0 \) and 324 is a perfect square (\( 18^2 \)), the roots are real, distinct, and rational. Since 324 is a perfect square, the roots are rational.
(iii) For the equation \( x^2 - 5x - 7 = 0 \):
We have \( a = 1 \), \( b = -5 \), and \( c = -7 \).
The discriminant is \( D = b^2 - 4ac \).
\( D = (-5)^2 - 4(1)(-7) \)
\( D = 25 + 28 \)
\( D = 53 \)
Since \( D > 0 \), the roots are real and distinct. However, since 53 is not a perfect square, the roots are irrational. Irrational roots often involve square roots that cannot be simplified to whole numbers or fractions.
(iv) For the equation \( a^2x^2 + abx = b^2 \), which can be written as \( a^2x^2 + abx - b^2 = 0 \):
We have \( a_{coeff} = a^2 \), \( b_{coeff} = ab \), and \( c_{coeff} = -b^2 \).
The discriminant is \( D = (ab)^2 - 4(a^2)(-b^2) \)
\( D = a^2b^2 + 4a^2b^2 \)
\( D = 5a^2b^2 \)
Since \( a \neq 0 \) and \( b \neq 0 \), \( 5a^2b^2 > 0 \). So, the roots are real and distinct. However, because 5 is not a perfect square, the roots are irrational. The presence of a non-square factor like 5 under the square root makes the roots irrational.
(v) For the equation \( 9a^2b^2x^2 - 48abcdx + 64c^2d^2 = 0 \):
We have \( a_{coeff} = 9a^2b^2 \), \( b_{coeff} = -48abcd \), and \( c_{coeff} = 64c^2d^2 \).
The discriminant is \( D = (-48abcd)^2 - 4(9a^2b^2)(64c^2d^2) \)
\( D = 2304a^2b^2c^2d^2 - 2304a^2b^2c^2d^2 \)
\( D = 0 \)
Since \( D = 0 \), the roots are real and equal. This form suggests the quadratic equation is a perfect square trinomial.
(vi) For the equation \( 4x^2 = 1 \), which can be written as \( 4x^2 - 1 = 0 \):
We have \( a = 4 \), \( b = 0 \), and \( c = -1 \).
The discriminant is \( D = b^2 - 4ac \).
\( D = (0)^2 - 4(4)(-1) \)
\( D = 0 + 16 \)
\( D = 16 \)
Since \( D > 0 \) and 16 is a perfect square (\( 4^2 \)), the roots are real, distinct, and rational. When the discriminant is a perfect square, the roots are always rational.
(vii) For the equation \( 64x^2 - 112x + 49 = 0 \):
We have \( a = 64 \), \( b = -112 \), and \( c = 49 \).
The discriminant is \( D = b^2 - 4ac \).
\( D = (-112)^2 - 4(64)(49) \)
\( D = 12544 - 12544 \)
\( D = 0 \)
Since \( D = 0 \), the roots are real and equal. This equation is a perfect square because \( (8x - 7)^2 = 64x^2 - 112x + 49 \).
(viii) For the equation \( 8x^2 + 5x + 1 = 0 \):
We have \( a = 8 \), \( b = 5 \), and \( c = 1 \).
The discriminant is \( D = b^2 - 4ac \).
\( D = (5)^2 - 4(8)(1) \)
\( D = 25 - 32 \)
\( D = -7 \)
Since \( D < 0 \), the roots are not real; they are imaginary. Complex numbers are needed to express imaginary roots.
In simple words: To know if a quadratic equation's solutions (roots) are real, equal, or imaginary without actually solving it, we look at the discriminant. If this number is positive, the roots are real and different. If it's zero, the roots are real and the same. If it's negative, there are no real roots.

🎯 Exam Tip: Pay close attention to the sign of the discriminant and whether it's a perfect square to accurately describe the nature of the roots. Also, ensure the equation is in standard form \( ax^2 + bx + c = 0 \) before identifying the coefficients.

Question 3. Find the value of 'k' so that the equation has equal roots (or coincident roots)
(i) \( 4x^2 + kx + 9 = 0 \)
(ii) \( kx^2 - 5x + k = 0 \)
(iii) \( 9x^2 + 3kx + 4 = 0 \)
(iv) \( x^2 + 7 (3 + 2k) - 2x (1 + 3k) = 0 \)
(v) \( (k - 12) x^2 + 2 (k - 12) x + 2 = 0 \)
Answer:
(i) For the equation \( 4x^2 + kx + 9 = 0 \):
The coefficients are \( a = 4 \), \( b = k \), and \( c = 9 \).
For equal roots, the discriminant \( D \) must be equal to 0.
\( D = b^2 - 4ac = 0 \)
\( k^2 - 4(4)(9) = 0 \)
\( k^2 - 144 = 0 \)
\( k^2 = 144 \)
\( k = \pm\sqrt{144} \)
\( k = \pm 12 \)
So, \( k = 12 \) or \( k = -12 \). The discriminant being zero means the parabola touches the x-axis at exactly one point.
(ii) For the equation \( kx^2 - 5x + k = 0 \):
The coefficients are \( a = k \), \( b = -5 \), and \( c = k \).
For equal roots, the discriminant \( D \) must be equal to 0.
\( D = b^2 - 4ac = 0 \)
\( (-5)^2 - 4(k)(k) = 0 \)
\( 25 - 4k^2 = 0 \)
\( 4k^2 = 25 \)
\( k^2 = \frac{25}{4} \)
\( k = \pm\sqrt{\frac{25}{4}} \)
\( k = \pm \frac{5}{2} \)
So, \( k = \frac{5}{2} \) or \( k = -\frac{5}{2} \). Remember to consider both positive and negative square roots when solving for k.
(iii) For the equation \( 9x^2 + 3kx + 4 = 0 \):
The coefficients are \( a = 9 \), \( b = 3k \), and \( c = 4 \).
For equal roots, the discriminant \( D \) must be equal to 0.
\( D = b^2 - 4ac = 0 \)
\( (3k)^2 - 4(9)(4) = 0 \)
\( 9k^2 - 144 = 0 \)
\( 9k^2 = 144 \)
\( k^2 = \frac{144}{9} \)
\( k^2 = 16 \)
\( k = \pm\sqrt{16} \)
\( k = \pm 4 \)
So, \( k = 4 \) or \( k = -4 \). Dividing the equation by 9 helps simplify the numbers before factoring.
(iv) For the equation \( x^2 + 7 (3 + 2k) - 2x (1 + 3k) = 0 \):
First, we rewrite the equation in standard form \( ax^2 + bx + c = 0 \):
\( x^2 - 2(1 + 3k)x + 7(3 + 2k) = 0 \)
The coefficients are \( a = 1 \), \( b = -2(1 + 3k) \), and \( c = 7(3 + 2k) \).
For equal roots, the discriminant \( D \) must be equal to 0.
\( D = b^2 - 4ac = 0 \)
\( [-2(1 + 3k)]^2 - 4(1)(7(3 + 2k)) = 0 \)
\( 4(1 + 6k + 9k^2) - 28(3 + 2k) = 0 \)
\( 4 + 24k + 36k^2 - 84 - 56k = 0 \)
\( 36k^2 - 32k - 80 = 0 \)
Divide the entire equation by 4:
\( 9k^2 - 8k - 20 = 0 \)
Now, we factor this quadratic equation for k:
\( 9k^2 - 18k + 10k - 20 = 0 \)
\( 9k(k - 2) + 10(k - 2) = 0 \)
\( (k - 2)(9k + 10) = 0 \)
This gives us two possible values for k:
\( k - 2 = 0 \implies k = 2 \)
\( 9k + 10 = 0 \implies 9k = -10 \implies k = -\frac{10}{9} \)
So, \( k = 2 \) or \( k = -\frac{10}{9} \). It is often helpful to factor out common numbers to simplify large coefficients.
(v) For the equation \( (k - 12) x^2 + 2 (k - 12) x + 2 = 0 \):
The coefficients are \( a = (k - 12) \), \( b = 2(k - 12) \), and \( c = 2 \).
For equal roots, the discriminant \( D \) must be equal to 0.
\( D = b^2 - 4ac = 0 \)
\( [2(k - 12)]^2 - 4(k - 12)(2) = 0 \)
\( 4(k - 12)^2 - 8(k - 12) = 0 \)
Factor out the common term \( 4(k - 12) \):
\( 4(k - 12)[(k - 12) - 2] = 0 \)
\( 4(k - 12)(k - 14) = 0 \)
This gives us two possible values for k:
\( k - 12 = 0 \implies k = 12 \)
\( k - 14 = 0 \implies k = 14 \)
So, \( k = 12 \) or \( k = 14 \). Factoring out the common term \( (k-12) \) makes solving the equation much easier.
In simple words: When a quadratic equation has roots that are the same, it means its discriminant (the part under the square root in the quadratic formula) is zero. We use this fact to create an equation for 'k' and then solve it. Remember to check for standard form first.

🎯 Exam Tip: When an equation has a common factor like \( (k-12) \), factoring it out early can simplify the algebra significantly and prevent errors.

Question 4. For what value of k, do the following quadratic equations have real roots.
(i) \( 2x^2 + 2x + k = 0 \)
(ii) \( kx^2 - 2x + 2 = 0 \)
(iii) \( 2x^2 - 10x + k = 0 \)
(iv) \( kx^2 + 8x - 2 = 0 \)
(v) \( 9x^2 - 24x + k = 0 \)
(vi) \( x^2 - 4x + k = 0 \)
Answer:
(i) For the equation \( 2x^2 + 2x + k = 0 \):
The coefficients are \( a = 2 \), \( b = 2 \), and \( c = k \).
For real roots, the discriminant \( D \) must be greater than or equal to 0 (\( D \ge 0 \)).
\( D = b^2 - 4ac \)
\( D = (2)^2 - 4(2)(k) \)
\( D = 4 - 8k \)
So, we need \( 4 - 8k \ge 0 \)
\( 4 \ge 8k \)
Divide by 4:
\( 1 \ge 2k \)
\( \frac{1}{2} \ge k \)
Thus, \( k \le \frac{1}{2} \). A smaller value of k helps keep the discriminant positive or zero.
(ii) For the equation \( kx^2 - 2x + 2 = 0 \):
The coefficients are \( a = k \), \( b = -2 \), and \( c = 2 \).
For real roots, the discriminant \( D \) must be greater than or equal to 0 (\( D \ge 0 \)).
\( D = b^2 - 4ac \)
\( D = (-2)^2 - 4(k)(2) \)
\( D = 4 - 8k \)
So, we need \( 4 - 8k \ge 0 \)
\( 4 \ge 8k \)
Divide by 4:
\( 1 \ge 2k \)
\( \frac{1}{2} \ge k \)
Thus, \( k \le \frac{1}{2} \). If k were zero, this would not be a quadratic equation anymore.
(iii) For the equation \( 2x^2 - 10x + k = 0 \):
The coefficients are \( a = 2 \), \( b = -10 \), and \( c = k \).
For real roots, the discriminant \( D \) must be greater than or equal to 0 (\( D \ge 0 \)).
\( D = b^2 - 4ac \)
\( D = (-10)^2 - 4(2)(k) \)
\( D = 100 - 8k \)
So, we need \( 100 - 8k \ge 0 \)
\( 100 \ge 8k \)
Divide by 4:
\( 25 \ge 2k \)
\( \frac{25}{2} \ge k \)
Thus, \( k \le \frac{25}{2} \). The value \( \frac{25}{2} \) marks the boundary between real and imaginary roots for this equation.
(iv) For the equation \( kx^2 + 8x - 2 = 0 \):
The coefficients are \( a = k \), \( b = 8 \), and \( c = -2 \).
For real roots, the discriminant \( D \) must be greater than or equal to 0 (\( D \ge 0 \)).
\( D = b^2 - 4ac \)
\( D = (8)^2 - 4(k)(-2) \)
\( D = 64 + 8k \)
So, we need \( 64 + 8k \ge 0 \)
\( 8k \ge -64 \)
Divide by 8:
\( k \ge -8 \)
Thus, \( k \ge -8 \). Quadratic equations can also have real roots even if they are not distinct.
(v) For the equation \( 9x^2 - 24x + k = 0 \):
The coefficients are \( a = 9 \), \( b = -24 \), and \( c = k \).
For real roots, the discriminant \( D \) must be greater than or equal to 0 (\( D \ge 0 \)).
\( D = b^2 - 4ac \)
\( D = (-24)^2 - 4(9)(k) \)
\( D = 576 - 36k \)
So, we need \( 576 - 36k \ge 0 \)
\( 576 \ge 36k \)
Divide by 36:
\( 16 \ge k \)
Thus, \( k \le 16 \). If k were greater than 16, the discriminant would become negative.
(vi) For the equation \( x^2 - 4x + k = 0 \):
The coefficients are \( a = 1 \), \( b = -4 \), and \( c = k \).
For real roots, the discriminant \( D \) must be greater than or equal to 0 (\( D \ge 0 \)).
\( D = b^2 - 4ac \)
\( D = (-4)^2 - 4(1)(k) \)
\( D = 16 - 4k \)
So, we need \( 16 - 4k \ge 0 \)
\( 16 \ge 4k \)
Divide by 4:
\( 4 \ge k \)
Thus, \( k \le 4 \). This means k can be any number from negative infinity up to 4.
In simple words: For a quadratic equation to have real solutions, its discriminant (the value \( b^2 - 4ac \)) must be zero or a positive number. If it's negative, the solutions are not real. We set up an inequality with the discriminant and solve it to find the possible values for 'k'.

🎯 Exam Tip: Remember that "real roots" includes both "real and distinct roots" (\( D > 0 \)) and "real and equal roots" (\( D = 0 \)), so the condition is \( D \ge 0 \).

Question 5. Find the value of x for which the given equation has equal roots. Also find the roots :
(i) \( 9x^2 - 24x + k = 0 \)
(ii) \( 2kx^2 - 40x + 25 = 0 \)
Answer:
(i) For the equation \( 9x^2 - 24x + k = 0 \):
The coefficients are \( a = 9 \), \( b = -24 \), and \( c = k \).
For equal roots, the discriminant \( D \) must be 0.
\( D = b^2 - 4ac = 0 \)
\( (-24)^2 - 4(9)(k) = 0 \)
\( 576 - 36k = 0 \)
\( 576 = 36k \)
\( k = \frac{576}{36} \)
\( k = 16 \)
Now, substitute \( k = 16 \) back into the original equation to find the roots:
\( 9x^2 - 24x + 16 = 0 \)
This equation is a perfect square trinomial:
\( (3x)^2 - 2(3x)(4) + (4)^2 = 0 \)
\( (3x - 4)^2 = 0 \)
To find x, take the square root of both sides:
\( 3x - 4 = 0 \)
\( 3x = 4 \)
\( x = \frac{4}{3} \)
Therefore, the value of k is 16, and the root is \( x = \frac{4}{3} \). When the discriminant is zero, the root found is a repeated root.
(ii) For the equation \( 2kx^2 - 40x + 25 = 0 \):
The coefficients are \( a = 2k \), \( b = -40 \), and \( c = 25 \).
For equal roots, the discriminant \( D \) must be 0.
\( D = b^2 - 4ac = 0 \)
\( (-40)^2 - 4(2k)(25) = 0 \)
\( 1600 - 200k = 0 \)
\( 1600 = 200k \)
\( k = \frac{1600}{200} \)
\( k = 8 \)
Now, substitute \( k = 8 \) back into the original equation to find the roots:
\( 2(8)x^2 - 40x + 25 = 0 \)
\( 16x^2 - 40x + 25 = 0 \)
This equation is a perfect square trinomial:
\( (4x)^2 - 2(4x)(5) + (5)^2 = 0 \)
\( (4x - 5)^2 = 0 \)
To find x, take the square root of both sides:
\( 4x - 5 = 0 \)
\( 4x = 5 \)
\( x = \frac{5}{4} \)
Therefore, the value of k is 8, and the root is \( x = \frac{5}{4} \). The question asks for the value of x, which represents the root of the equation.
In simple words: If an equation has equal roots, it means its discriminant is zero. First, use this rule to find the unknown value 'k'. Then, put 'k' back into the original equation and solve for 'x' to find the actual root.

🎯 Exam Tip: Recognizing a perfect square trinomial can save time when finding roots after determining \( D = 0 \). Always remember to find both k and x if the question asks for both.

Question 6. Find the values of k for which the given quadratic equations has real and distinct roots :
(i) \( kx^2 + 2x + 1 = 0 \)
(ii) \( kx^2 + 6x + 1 = 0 \)
Answer:
(i) For the equation \( kx^2 + 2x + 1 = 0 \):
The coefficients are \( a = k \), \( b = 2 \), and \( c = 1 \).
For real and distinct roots, the discriminant \( D \) must be strictly greater than 0 (\( D > 0 \)).
\( D = b^2 - 4ac \)
\( D = (2)^2 - 4(k)(1) \)
\( D = 4 - 4k \)
So, we need \( 4 - 4k > 0 \)
\( 4 > 4k \)
Divide by 4:
\( 1 > k \)
Thus, \( k < 1 \). Remember that k cannot be zero, otherwise it would not be a quadratic equation.
(ii) For the equation \( kx^2 + 6x + 1 = 0 \):
The coefficients are \( a = k \), \( b = 6 \), and \( c = 1 \).
For real and distinct roots, the discriminant \( D \) must be strictly greater than 0 (\( D > 0 \)).
\( D = b^2 - 4ac \)
\( D = (6)^2 - 4(k)(1) \)
\( D = 36 - 4k \)
So, we need \( 36 - 4k > 0 \)
\( 36 > 4k \)
Divide by 4:
\( 9 > k \)
Thus, \( k < 9 \). The condition \( k < 9 \) means k can be any number less than 9, but not 0 to keep it a quadratic.
In simple words: To have two different real solutions, the discriminant of the quadratic equation must be a positive number (not zero, not negative). We calculate the discriminant and set it to be greater than zero, then solve for 'k'.

🎯 Exam Tip: Ensure you use the strict inequality \( D > 0 \) for "real and distinct roots" and not \( D \ge 0 \), which includes equal roots.

Question 7. If – 5 is a roots of the quadratic equation \( 2x^2 + px - 15 = 0 \) and the quadratic equation \( p(x^2 + x) + k = 0 \) has equal roots, find the value of k.
Answer:
First, we need to find the value of 'p'. We are given that \( x = -5 \) is a root of the equation \( 2x^2 + px - 15 = 0 \).
This means if we substitute \( x = -5 \) into the equation, it will be true:
\( 2(-5)^2 + p(-5) - 15 = 0 \)
\( 2(25) - 5p - 15 = 0 \)
\( 50 - 5p - 15 = 0 \)
\( 35 - 5p = 0 \)
\( 5p = 35 \)
\( p = \frac{35}{5} \)
\( p = 7 \)
Now that we have \( p = 7 \), we can use the second quadratic equation: \( p(x^2 + x) + k = 0 \).
Substitute \( p = 7 \) into this equation:
\( 7(x^2 + x) + k = 0 \)
Expand this to the standard quadratic form \( ax^2 + bx + c = 0 \):
\( 7x^2 + 7x + k = 0 \)
We are told this equation has equal roots. For equal roots, the discriminant \( D \) must be 0.
The coefficients are \( a = 7 \), \( b = 7 \), and \( c = k \).
\( D = b^2 - 4ac = 0 \)
\( (7)^2 - 4(7)(k) = 0 \)
\( 49 - 28k = 0 \)
\( 49 = 28k \)
\( k = \frac{49}{28} \)
Simplify the fraction by dividing both numerator and denominator by 7:
\( k = \frac{7}{4} \)
Thus, the value of k is \( \frac{7}{4} \). This problem shows how properties of one quadratic equation can help solve for unknowns in another.
In simple words: First, use the given root to find the value of 'p' in the first equation. Then, use this 'p' value in the second equation. Since the second equation has equal roots, its discriminant must be zero. Use this rule to solve for 'k'.

🎯 Exam Tip: Break down complex problems into smaller, manageable steps. If one part of the problem gives information about another, ensure you use that information correctly in the subsequent steps.