CBSE Class 10 Maths HOTs Quadratic Equations Set 16

Question. If \( p, q \) and \( r \) are rational numbers and \( p \neq q \neq r \), then roots of the equation \( (p^2 - q^2)x^2 - (q^2 - r^2)x + (r^2 - p^2) = 0 \) are
(a) \( \frac{p}{q}, \frac{r}{p} \)
(b) \( \frac{p^2}{q^2}, \frac{r^2}{q^2} \)
(c) \( 1, \frac{r^2 - p^2}{p^2 - q^2} \)
(d) \( -1, \frac{p^2 - q^2}{p^2 - r^2} \)
Answer: (d) \( -1, \frac{p^2 - q^2}{p^2 - r^2} \)
Putting \( x = -1 \), we have
\( (p^2 - q^2)(-1)^2 - (q^2 - r^2)(-1) + (r^2 - p^2) \)
\( \implies p^2 - q^2 + q^2 - r^2 + r^2 - p^2 = 0 \)
\( \therefore x = -1 \) is one root. Only option (d) has one root \( -1 \).

Question. If \( \alpha, \beta \) are roots of the equation \( x^2 + 5x + 5 = 0 \), then equation whose roots are \( \alpha + 1 \) and \( \beta + 1 \) is
(a) \( x^2 + 5x - 5 = 0 \)
(b) \( x^2 + 3x + 5 = 0 \)
(c) \( x^2 + 3x + 1 = 0 \)
(d) None of the options
Answer: (c) \( x^2 + 3x + 1 = 0 \)
\( \alpha + \beta = -5, \alpha\beta = 5. \) Required equation is
\( x^2 - (\alpha + 1 + \beta + 1)x + (\alpha + 1)(\beta + 1) = 0 \)
\( \implies x^2 - (\alpha + \beta + 2)x + (\alpha\beta + \alpha + \beta + 1) = 0 \)
\( \implies x^2 - (-5 + 2)x + (5 - 5 + 1) = 0 \)
\( \implies x^2 + 3x + 1 = 0 \)

Question. If \( \alpha, \beta \) are roots of \( x^2 + 5x + a = 0 \) and \( 2\alpha + 5\beta = -1 \), then the value of \( a \) is
(a) 24
(b) -24
(c) \( \frac{1}{24} \)
(d) -10
Answer: (b) -24
Here \( \alpha + \beta = -5 \) ...(i)
and \( 2\alpha + 5\beta = -1 \) ...(ii)
Multiplying (i) by 2, we get
\( \implies 2\alpha + 2\beta = -10 \) ...(iii)
Solving (ii) and (iii), we get \( \alpha = -8, \beta = 3 \)
Now \( \alpha\beta = \frac{a}{1} \implies a = -24 \)

Question. \( \alpha, \beta \) are roots of the equation \( (a + 1)x^2 + (2a + 3)x + (3a + 4) = 0 \). If \( \alpha \cdot \beta = 2 \), then find the value of \( \alpha + \beta \)
Answer: \( \because \alpha\beta = \frac{3a + 4}{a + 1} \)
\( \implies 2 = \frac{3a + 4}{a + 1} \)
\( \implies 2a + 2 = 3a + 4 \implies a = -2 \)
and \( \alpha + \beta = -\left( \frac{2a + 3}{a + 1} \right) \)
\( \implies \alpha + \beta = -\left( \frac{2(-2) + 3}{-2 + 1} \right) = -\left( \frac{-1}{-1} \right) = -1 \)

Question. For what value of \( k \), the roots of the equation \( x^2 + 4x + k = 0 \) are real? 
Answer: For real roots; \( b^2 - 4ac \geq 0 \)
\( \therefore (4)^2 - 4 \times 1 \times k \geq 0 \)
\( \implies 16 - 4k \geq 0 \implies 4k \leq 16 \implies k \leq 4. \)

Question. Write the nature of roots of quadratic equation \( 4x^2 + 4\sqrt{3}x + 3 = 0. \)
Answer: Given equation is \( 4x^2 + 4\sqrt{3}x + 3 = 0, \)
Here \( a = 4, b = 4\sqrt{3}, c = 3 \)
\( D = b^2 - 4ac = 48 - 48 = 0 \)
As \( D = 0, \) the equation has real and equal roots.

Question. Write the nature of roots of the quadratic equation \( 9x^2 - 6x - 2 = 0. \)
Answer: Given quadratic equation is \( 9x^2 - 6x - 2 = 0 \)
Here \( a = 9, b = -6, c = -2 \)
\( D = b^2 - 4ac \)
\( \implies D = (-6)^2 - 4 \times 9 \times (-2) = 36 + 72 = 108 > 0 \)
\( \therefore \) Given quadratic has two unequal real roots.

Question. Write the nature of roots of quadratic equation : \( 4x^2 + 6x + 3 = 0 \)
Answer: Given quadratic equation is \( 4x^2 + 6x + 3 = 0. \)
Here, \( a = 4, b = 6, c = 3 \)
\( D = b^2 - 4ac \)
\( \implies D = (6)^2 - 4 \times 4 \times 3 = 36 - 48 = -12 < 0 \)
\( \therefore \) given quadratic equation has no real roots.

Question. If arithmetic mean of two numbers \( a \) and \( b \) is 8 and \( ab = 9 \), find a quadratic equation whose roots are \( a \) and \( b \).
Answer: Here \( \frac{a + b}{2} = 8 \implies a + b = 16 \) and \( ab = 9 \)
Now, quadratic equation whose roots are \( a \) and \( b \) is
\( x^2 - (a + b)x + ab = 0 \implies x^2 - 16x + 9 = 0 \)

Question. If \( 2x^2 - (2 + k)x + k = 0 \) where \( k \) is a real number, find the roots of the equation.
Answer: Given quadratic equation is \( 2x^2 - (2 + k)x + k = 0. \)
Here, \( a = 2, b = -(2 + k), c = k \)
Now, \( a + b + c = 2 + [-(2 + k)] + k = 0 \)
\( \therefore \) roots are \( 1 \) and \( \frac{k}{2} \).
(If \( a + b + c = 0 \), then roots of the quadratic equation are \( 1 \) and \( \frac{c}{a} \))

Question. Find the value of \( k \) for which the equation \( x^2 + k(2x + k - 1) = 0 \) has real and equal roots. 
Answer: \( x^2 + k(2x + k - 1) = 0 \)
\( \implies x^2 + 2kx + k^2 - k = 0 \)
Here, \( a = 1, b = 2k, c = k^2 - k \)
Since, the roots are real and equal
\( \therefore D = b^2 - 4ac = 0 \)
\( \implies (2k)^2 - 4 \times 1 \times (k^2 - k) = 0 \)
\( \implies 4k^2 - 4k^2 + 4k = 0 \)
\( \implies 4k = 0 \)
\( \therefore k = \frac{0}{4} = 0 \)

Question. Find the value of \( p \), for which one root of the quadratic equation \( px^2 - 14x + 8 = 0 \) is 6 times the other. 
Answer: The given quadratic equation is \( px^2 - 14x + 8 = 0 \)
Here \( a = p, b = -14 \) and \( c = 8 \)
Let \( \alpha, \beta \) be the two roots of this equation such that \( \alpha = 6\beta \)
Now, sum of roots = \( \frac{-b}{a} \)
\( \implies \alpha + \beta = \frac{-b}{a} \)
\( \implies 6\beta + \beta = -\left(\frac{-14}{p}\right) \)
\( \implies 7\beta = \frac{14}{p} \) or \( \beta = \frac{2}{p} \) ...(i)
Also product of roots = \( \frac{c}{a} \)
\( \implies \alpha \cdot \beta = \frac{c}{a} \)
\( \implies 6\beta \cdot \beta = \frac{8}{p} \)
\( \implies \beta^2 = \frac{8}{6p} = \frac{4}{3p} \) ...(ii)
Solving equations (i) and (ii) to eliminate \( \beta \) we get
\( \left(\frac{2}{p}\right)^2 = \frac{4}{3p} \implies \frac{4}{p^2} = \frac{4}{3p} \)
\( \implies p^2 - 3p = 0 \)
\( \implies p(p - 3) = 0 \implies p = 0, 3 \)
If \( p = 0 \), then the given equation will not remain quadratic.
So, \( p = 0 \) is rejected. Hence, the value of \( p \) is 3.

Question. Find the value of \( p \) so that the quadratic equation \( px(x - 3) + 9 = 0 \) has two equal roots. 
Answer: \( px(x - 3) + 9 = 0 \)
\( \implies px^2 - 3px + 9 = 0. \)
Here, \( a = p, b = -3p, c = 9 \)
For equal roots \( D = 0 \)
\( \implies D = b^2 - 4ac = 0 \)
\( \implies (-3p)^2 - 4 \times p \times 9 = 0 \)
\( \implies 9p^2 - 36p = 0 \)
\( \implies 9p(p - 4) = 0 \)
\( \implies 9p = 0 \) or \( p - 4 = 0 \)
\( \implies p = 0 \) or \( p = 4 \)
but \( p \neq 0 \) [\( \because \) In quadratic equation \( a \neq 0 \)]
\( \therefore p = 4 \)

Question. Is the following situation possible? The sum of ages of a mother and her daughter is 25 years. Five years ago the product of their ages was 58.
Answer: Let age of mother be \( x \) years
\( \therefore \) Age of the daughter be \( (25 - x) \) years
Five years ago,
Mother’s age = \( (x - 5) \) years
and daughter’s age = \( (25 - x - 5) = (20 - x) \) years
ATQ \( (x - 5)(20 - x) = 58 \)
\( \implies 20x - x^2 - 100 + 5x = 58 \)
\( \implies x^2 - 25x + 158 = 0. \)
So \( a = 1, b = -25, c = 158 \)
\( D = b^2 - 4ac = (-25)^2 - 4 \times 1 \times 158 = 625 - 632 = -7 < 0 \)
\( \therefore \) No real solution of the equation is possible.
Hence, the given situation is not possible.

Question. Find the values of \( a \) and \( b \), if the sum and the product of the roots of the equation \( 4ax^2 + 4bx + 3 = 0 \) are 1/2 and 3/16 respectively.
Answer: The given equation is \( 4ax^2 + 4bx + 3 = 0. \) Let \( \alpha \) and \( \beta \) are the roots of equation
Here \( A = 4a, B = 4b, C = 3. \)
Sum of roots = \( -\frac{B}{A} = -\frac{4b}{4a} = -\frac{b}{a} \)
\( \therefore -\frac{b}{a} = \frac{1}{2} \) [Given]
\( \implies -2b = a \) ... (i)
Product of roots = \( \frac{C}{A} \)
\( \implies \frac{3}{16} = \frac{3}{4a} \)
\( \implies 4a = 16 \implies a = 4 \)
Putting in equation (i), we get \( b = -2. \)
Hence, \( a = 4 \) and \( b = -2 \)

Question. For what value of \( k \), the given equation \( (4 - k)x^2 + (2k + 4)x + (8k + 1) = 0 \) is a perfect square?
Answer: \( (4 - k)x^2 + (2k + 4)x + (8k + 1) = 0 \)
\( a = 4 - k, b = 2k + 4, c = 8k + 1 \)
\( D = b^2 - 4ac \)
\( \implies D = (2k+4)^2 - 4(4 - k)(8k + 1) \)
\( = 4k^2 + 16k + 16 - 4(32k + 4 - 8k^2 - k) \)
\( = 4k^2 + 16k + 16 - 128k - 16 + 32k^2 + 4k \)
\( = 36k^2 - 108k \)
For perfect square, \( D = 0 \)
\( \implies 36k^2 - 108k = 0 \)
\( \implies 36k(k - 3) = 0 \implies k = 0, k = 3 \)

Question. Find the value of \( k \) for which the equation \( x^2 + (1 - 2k)x + k^2 - 1 = 0 \) has only one real value of \( x \) as its solution.
Answer: Given quadratic equation is \( x^2 + (1 - 2k)x + k^2 - 1 = 0 \)
Here \( a = 1, b = 1 - 2k, c = k^2 - 1 \)
\( D = b^2 - 4ac \)
\( = (1 - 2k)^2 - 4 \times 1 \times (k^2 - 1) \)
\( = 1 + 4k^2 - 4k - 4k^2 + 4 = 5 - 4k \)
\( \because \) there is only one real value of \( x \) of its solution
\( \implies \) both the roots of the equation are equal
\( \implies D = 0 \implies 5 - 4k = 0 \implies k = \frac{5}{4} \)

Question. If the equation \( (1 + m^2)x^2 + 2mcx + c^2 - a^2 = 0 \) has equal roots, then show that \( c^2 = a^2(1 + m^2). \)
Answer: The given equation is \( (1 + m^2)x^2 + 2mcx + c^2 - a^2 = 0 \)
Here \( A = (1 + m^2), B = 2mc, C = c^2 - a^2 \)
\( \because \) This equation has equal roots
\( \therefore D = 0 \)
\( \implies B^2 - 4AC = 0 \)
\( \implies (2mc)^2 - 4(1 + m^2)(c^2 - a^2) = 0 \)
\( \implies 4m^2c^2 - 4(c^2 - a^2 + m^2c^2 - m^2a^2) = 0 \)
\( \implies 4m^2c^2 - 4c^2 + 4a^2 - 4m^2c^2 + 4m^2a^2 = 0 \)
\( \implies -4c^2 + 4a^2 + 4m^2a^2 = 0 \)
\( \implies 4c^2 - 4a^2 - 4m^2a^2 = 0 \)
\( \implies c^2 - a^2 - m^2a^2 = 0 \)
\( \implies c^2 = a^2 + m^2a^2 \)
\( \implies c^2 = a^2(1 + m^2) \)

Question. If \( ad \neq bc \), then prove that the equation \( (a^2 + b^2)x^2 + 2(ac + bd)x + (c^2 + d^2) = 0 \) has no real roots. 
Answer: The given equation is \( (a^2 + b^2)x^2 + 2(ac + bd)x + (c^2 + d^2) = 0 \)
Here \( A = (a^2 + b^2), B = 2(ac + bd), C = (c^2 + d^2) \)
To find the nature of its roots, we find the discriminant.
\( D = B^2 - 4AC \)
\( = [2(ac + bd)]^2 - 4(a^2 + b^2)(c^2 + d^2) \)
\( = 4(a^2c^2 + b^2d^2 + 2abcd) - 4(a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2) \)
\( = 4a^2c^2 + 4b^2d^2 + 8abcd - 4a^2c^2 - 4a^2d^2 - 4b^2c^2 - 4b^2d^2 \)
\( = -4a^2d^2 - 4b^2c^2 + 8abcd \)
\( = -4(a^2d^2 + b^2c^2 - 2abcd) = -4(ad - bc)^2 \)
It is given that \( ad \neq bc \)
\( \implies ad - bc \neq 0 \)
\( \implies (ad - bc)^2 > 0 \) for all values of \( a, b, c \) and \( d \)
\( \implies -4(ad - bc)^2 < 0 \)
\( \implies D < 0 \)
Hence, the given equation has no real roots.

Question. Show that if the roots of the following quadratic equation are equal, then \( ad = bc \)
\( x^2(a^2 + b^2) + 2(ac + bd)x + (c^2 + d^2) = 0 \)
Answer: Given equation is \( x^2(a^2 + b^2) + 2(ac + bd)x + (c^2 + d^2) = 0 \)
Roots are equal, \( \therefore B^2 - 4AC = 0 \)
\( [2(ac + bd)]^2 - 4[a^2 + b^2][c^2 + d^2] = 0 \)
\( 4a^2c^2 + 4b^2d^2 + 8acbd - 4a^2c^2 - 4a^2d^2 - 4b^2c^2 - 4b^2d^2 = 0 \)
\( 8acbd - 4a^2d^2 - 4b^2c^2 = 0 \)
\( \implies -4(ad - bc)^2 = 0 \)
\( \implies ad - bc = 0 \implies ad = bc \)

Question. If \( a \) and \( b \) are roots of the equation \( 2x^2 + 7x + 5 = 0 \) then write a quadratic equation whose roots are \( 2a + 3 \) and \( 2b + 3 \).
Answer: Here given quadratic equation is \( 2x^2 + 7x + 5 = 0 \)
\( \because a \) and \( b \) are roots
\( \therefore a + b = -\frac{7}{2} \) ...(i)
and \( a \cdot b = \frac{5}{2} \) ...(ii)
Now, quadratic equation whose roots are \( 2a + 3 \) and \( 2b + 3 \) is
\( x^2 - [2a + 3 + 2b + 3]x + (2a + 3)(2b + 3) = 0 \)
\( \implies x^2 - [2(a + b) + 6]x + [4ab + 6(a + b) + 9] = 0 \)
\( \implies x^2 - [2(-\frac{7}{2}) + 6]x + [4(\frac{5}{2}) + 6(-\frac{7}{2}) + 9] = 0 \) [using eq. (i) and (ii)]
\( \implies x^2 + x - 2 = 0 \)

Question. If difference of the roots of the equation \( x^2 - 7x + 2k = 0 \) is 1 then find the value of \( k \).
Answer: Let \( \alpha \) and \( \beta (\alpha > \beta) \) are the roots of the equation \( x^2 - 7x + 2k = 0 \)
\( \implies \alpha + \beta = -(\frac{-7}{1}) = 7 \) ...(i)
and \( \alpha\beta = 2k \) ...(ii)
ATQ \( \alpha - \beta = 1 \)
\( \implies (\alpha - \beta)^2 = 1 \)
\( \implies (\alpha + \beta)^2 - 4\alpha\beta = 1 \) [using \( (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta \)]
\( \implies (7)^2 - 4 \times 2k = 1 \) [using (i) and (ii)]
\( \implies 49 - 8k = 1 \implies -8k = -48 \implies k = 6 \)

Question. Find the positive value of \( k \), for which the equation \( x^2 + kx + 64 = 0 \) and \( x^2 - 8x + k = 0 \) will both have real roots. 
Answer: If the equation \( x^2 + kx + 64 = 0 \) has real roots, then \( D \geq 0. \)
\( \implies k^2 - 4 \times 64 \geq 0 \)
\( \implies k^2 \geq 256 \)
\( \implies k^2 \geq (16)^2 \implies k \geq 16 \) [\( \because k > 0 \)] ...(i)
If the equation \( x^2 - 8x + k = 0 \) has real roots, then \( D \geq 0 \)
\( \implies 64 - 4k \geq 0 \implies 4k \leq 64 \implies k \leq 16 \) ...(ii)
From (i) and (ii), we get \( k = 16. \)

Question. If \( -4 \) is a root of the quadratic equation \( x^2 + px - 4 = 0 \) and the quadratic equation \( x^2 + px + k = 0 \) has equal roots, find the value of \( k \).
Answer: \( -4 \) is a root of quadratic equation \( x^2 + px - 4 = 0 \)
\( \therefore (-4)^2 + p(-4) - 4 = 0 \)
\( \implies 16 - 4p - 4 = 0 \implies p = 3 \)
Putting \( p = 3 \) in equation \( x^2 + px + k = 0, \) we get \( x^2 + 3x + k = 0. \)
Equation has equal roots \( D = 0 \)
\( \implies b^2 - 4ac = 0 \)
\( \implies (3)^2 - 4 \times 1 \times k = 0 \)
\( \implies -4k = -9 \implies k = \frac{9}{4} \)

Question. Write all the values of \( p \) for which the quadratic equation \( x^2 + px + 16 = 0 \) has equal roots. Find the roots of the equation so obtained. 
Answer: \( x^2 + px + 16 = 0 \)
\( \therefore D = b^2 - 4ac = p^2 - 4 \times 1 \times 16 = p^2 - 64 \)
For equal roots, \( D = 0 \implies p^2 - 64 = 0 \implies p = \pm 8 \)
For \( p = 8, \) we have \( x^2 + 8x + 16 = 0 \)
\( x^2 + 4x + 4x + 16 = 0 \implies x(x + 4) + 4(x + 4) = 0 \implies x = -4, -4 \)
For \( p = -8, \) we have \( x^2 - 8x + 16 = 0 \)
\( x^2 - 4x - 4x + 16 = 0 \implies x(x - 4) - 4(x - 4) = 0 \implies x = 4, 4 \)

Question. If the roots of the equation \( px^2 + qx + r = 0 \) are real and of the form \( \frac{\alpha}{\alpha + 1} \) and \( \frac{\alpha + 1}{\alpha} \), then show that \( (p + q + r)^2 = 4p^2 + q^2 + 4pq. \)
Answer: Roots of the equation \( px^2 + qx + r = 0 \) are of the form \( \frac{\alpha}{\alpha + 1} \) and \( \frac{\alpha + 1}{\alpha} \)
\( \therefore \) Product of roots = \( \frac{\alpha}{\alpha + 1} \times \frac{\alpha + 1}{\alpha} = \frac{r}{p} \implies \frac{r}{p} = 1 \implies r = p \)
Now, \( (p + q + r)^2 = (p + q + p)^2 = (2p + q)^2 = 4p^2 + q^2 + 4pq \)

Question. Prove that both the roots of equation: \( (x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0 \) are equal if \( a = b = c. \)
Answer: \( (x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0 \)
\( \implies x^2 - (a + b)x + ab + x^2 - (b + c)x + bc + x^2 - (c + a)x + ac = 0 \)
\( \implies 3x^2 - 2(a + b + c)x + ab + bc + ac = 0 \)
\( D = [-2(a + b + c)]^2 - 4 \times 3 \times (ab + bc + ac) \)
\( = 4(a^2 + b^2 + c^2 + 2ab + 2bc + 2ac) - 12ab - 12bc - 12ac \)
\( \implies D = 4a^2 + 4b^2 + 4c^2 - 4ab - 4bc - 4ac \)
\( \implies D = 2[2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ac] \)
\( D = 2[(a^2 + b^2 - 2ab) + (b^2 + c^2 - 2bc) + (a^2 + c^2 - 2ac)] \)
\( D = 2[(a - b)^2 + (b - c)^2 + (c - a)^2] \)
For equal roots \( D = 0 \)
\( \implies 2[(a - b)^2 + (b - c)^2 + (c - a)^2] = 0 \)
\( \implies (a - b)^2 + (b - c)^2 + (c - a)^2 = 0 \)
\( \implies (a - b)^2 = 0, (b - c)^2 = 0, (c - a)^2 = 0 \)
(If sum of perfect squares is equal to zero, then each of them is zero)
\( \implies a = b, b = c, c = a \implies a = b = c. \)

Question. If sum of the two roots of the equation \( \frac{1}{x + a} + \frac{1}{x + b} = \frac{1}{c} \) is zero, then show that product of the two roots = \( -\frac{(a^2 + b^2)}{2} \)
Answer: \( \frac{1}{x + a} + \frac{1}{x + b} = \frac{1}{c} \)
\( \implies \frac{(x + b) + (x + a)}{(x + a)(x + b)} = \frac{1}{c} \)
\( \implies 2cx + bc + ca = x^2 + (a + b)x + ab \)
\( \implies x^2 + (a + b - 2c)x + ab - bc - ca = 0 \)
\( \because \) Sum of the roots = 0
\( \implies -\left(\frac{a + b - 2c}{1}\right) = 0 \implies a + b = 2c \implies c = \frac{a + b}{2} \) ...(i)
Product of roots = \( \frac{ab - bc - ca}{1} = ab - c(a + b) = ab - \left(\frac{a + b}{2}\right)(a + b) \) [using (i)]
\( = \frac{2ab - (a + b)^2}{2} = \frac{2ab - (a^2 + b^2 + 2ab)}{2} = -\frac{(a^2 + b^2)}{2} \)
Hence proved.

Question. For what value of \( k \), equations \( 2x^2 + kx - 5 = 0 \) and \( x^2 - 3x - 4 = 0 \) may have one root common.
Answer: Let \( y \) is a common root.
\( \implies 2y^2 + ky - 5 = 0 \) ...(i)
and \( y^2 - 3y - 4 = 0 \) ...(ii)
Multiplying eq. (ii) with 2, we get \( 2y^2 - 6y - 8 = 0 \) ...(iii)
eq. (i) - eq. (iii), we get \( (k + 6)y + 3 = 0 \implies y = -\frac{3}{k + 6} \)
Substituting \( y = -\frac{3}{k + 6} \) in eq. (ii), we get
\( \left(-\frac{3}{k + 6}\right)^2 - 3\left(-\frac{3}{k + 6}\right) - 4 = 0 \)
\( \implies \frac{9}{(k + 6)^2} + \frac{9}{k + 6} - 4 = 0 \)
\( \implies \frac{9 + 9(k + 6) - 4(k + 6)^2}{(k + 6)^2} = 0 \)
\( 9 + 9k + 54 - 4k^2 - 144 - 48k = 0 \implies 4k^2 + 39k + 81 = 0 \)
\( \implies 4k^2 + 27k + 12k + 81 = 0 \implies k(4k + 27) + 3(4k + 27) = 0 \)
\( \implies (4k + 27)(k + 3) = 0 \implies k = -\frac{27}{4} \) or \( k = -3 \)

Question. If the difference of the roots of the equation \( x^2 - bx + c = 0 \) be 1, then
(a) \( b^2 - 4c + 1 = 0 \)
(b) \( b^2 + 4c = 0 \)
(c) \( b^2 - 4c - 1 = 0 \)
(d) \( b^2 - 4c = 0 \)
Answer: (c) \( b^2 - 4c - 1 = 0 \)

Question. If the roots of \( ax^2 + bx + c = 0 \) are equal in magnitude but opposite in sign, then
(a) \( a = 0 \)
(b) \( b = 0 \)
(c) \( c = 0 \)
(d) None of the options
Answer: (b) \( b = 0 \)

Question. If \( \alpha + \beta = 4 \) and \( \alpha^3 + \beta^3 = 44 \), then \( \alpha, \beta \) are the roots of the equation
(a) \( 2x^2 - 7x - 7 = 0 \)
(b) \( 3x^2 + 8x + 12 = 0 \)
(c) \( 3x^2 - 12x + 5 = 0 \)
(d) None of the options
Answer: (c) \( 3x^2 - 12x + 5 = 0 \)

Question. If the roots of equation \( 3x^2 + 2x + (p + 2)(p - 1) = 0 \) are of opposite sign then which of the following cannot be the value of \( p \)?
(a) 0
(b) -1
(c) \( \frac{1}{2} \)
(d) -3
Answer: (d) -3

Question. The value of \( k \) for which the equation \( x^2 + 2(k + 1)x + k^2 = 0 \) has equal roots is
(a) -1
(b) \( -\frac{1}{2} \)
(c) 1
(d) None of the options
Answer: (b) \( -\frac{1}{2} \)

Question. If the equation \( x^2 - (2 + m)x + (-m^2 - 4m - 4) = 0 \) has coincident roots, then
(a) \( m = 0, m = 1 \)
(b) \( m = 2, m = 2 \)
(c) \( m = -2, m = -2 \)
(d) \( m = 6, m = 1 \)
Answer: (c) \( m = -2, m = -2 \)

Question. For what value of \( k \) does \( (k - 12)x^2 + 2(k - 12)x + 2 = 0 \) have equal roots?
Answer: \( D = 0 \implies [2(k-12)]^2 - 4(k-12)(2) = 0 \implies 4(k-12)^2 - 8(k-12) = 0 \implies 4(k-12)[(k-12) - 2] = 0 \implies k = 12 \) or \( k = 14. \) Since \( k \neq 12, k = 14. \)

Question. If two roots of \( 2x^2 + bx + c = 0 \) are reciprocal of each other then find the value of \( c \).
Answer: Product of roots = \( \alpha \cdot \frac{1}{\alpha} = \frac{c}{2} \implies 1 = \frac{c}{2} \implies c = 2. \)

Question. Without solving, find the nature of the roots of the following quadratic equations.
(i) \( 5x^2 - 3x - 4 = 0 \)
(ii) \( 4x^2 - 12x + 9 = 0 \)
Answer: (i) \( D = (-3)^2 - 4(5)(-4) = 9 + 80 = 89 > 0 \). Roots are real and unequal.
(ii) \( D = (-12)^2 - 4(4)(9) = 144 - 144 = 0 \). Roots are real and equal.

Question. The sum of the squares of three consecutive positive integers is 50. Find the integers.
Answer: Let integers be \( x, x+1, x+2. \) \( x^2 + (x+1)^2 + (x+2)^2 = 50 \implies 3x^2 + 6x + 5 = 50 \implies 3x^2 + 6x - 45 = 0 \implies x^2 + 2x - 15 = 0 \implies (x+5)(x-3) = 0 \implies x = 3. \) Integers are 3, 4, 5.