CBSE Class 10 Maths HOTs Quadratic Equations Set 17

Question. Determine whether the given quadratic equations have real roots, if so, find the roots:
(i) \( 2x^2 + 5\sqrt{3}x + 6 = 0 \)
(ii) \( 6x^2 + x - 2 = 0 \)
Answer: (i) \( D = (5\sqrt{3})^2 - 4(2)(6) = 75 - 48 = 27 \)
Since \( D > 0 \), roots are real.
\( x = \frac{-5\sqrt{3} \pm \sqrt{27}}{2(2)} = \frac{-5\sqrt{3} \pm 3\sqrt{3}}{4} \)
\( \implies x = \frac{-2\sqrt{3}}{4} = -\frac{\sqrt{3}}{2} \) and \( x = \frac{-8\sqrt{3}}{4} = -2\sqrt{3} \)
(ii) \( D = (1)^2 - 4(6)(-2) = 1 + 48 = 49 \)
Since \( D > 0 \), roots are real.
\( x = \frac{-1 \pm \sqrt{49}}{2(6)} = \frac{-1 \pm 7}{12} \)
\( \implies x = \frac{6}{12} = \frac{1}{2} \) and \( x = \frac{-8}{12} = -\frac{2}{3} \)

Question. Determine the positive value of \( k \) for which the equation \( 2x^2 + kx + 8 = 0 \) have real roots.
Answer: For real roots, \( D \geq 0 \).
\( k^2 - 4(2)(8) \geq 0 \)
\( \implies k^2 - 64 \geq 0 \)
\( \implies k^2 \geq 64 \)
\( \implies k \geq 8 \) or \( k \leq -8 \)
The positive value is \( k \geq 8 \).

Question. If the roots of the equation \( (b - c)x^2 + (c - a)x + (a - b) = 0 \) are equal, then prove that \( 2b = a + c \). 
Answer: Sum of coefficients \( = (b - c) + (c - a) + (a - b) = 0 \).
If the sum of coefficients is zero, one root is 1. Since roots are equal, both roots are 1.
Product of roots \( = 1 \cdot 1 = \frac{a - b}{b - c} \)
\( \implies 1 = \frac{a - b}{b - c} \)
\( \implies b - c = a - b \)
\( \implies 2b = a + c \)
Hence proved.

Question. Find the value of \( k \) so that the quadratic equation \( x^2 - 2x(1 + 3k) + 7(3 + 2k) = 0 \) has equal roots.
Answer: For equal roots, \( D = 0 \).
\( [-2(1 + 3k)]^2 - 4(1)[7(3 + 2k)] = 0 \)
\( \implies 4(1 + 9k^2 + 6k) - 28(3 + 2k) = 0 \)
\( \implies 4 + 36k^2 + 24k - 84 - 56k = 0 \)
\( \implies 36k^2 - 32k - 80 = 0 \)
\( \implies 9k^2 - 8k - 20 = 0 \)
\( \implies 9k^2 - 18k + 10k - 20 = 0 \)
\( \implies 9k(k - 2) + 10(k - 2) = 0 \)
\( \implies (9k + 10)(k - 2) = 0 \)
\( \implies k = 2 \) or \( k = -\frac{10}{9} \)

Question. If \( \sin\alpha \) and \( \cos\alpha \) are the roots of the equation \( ax^2 + bx + c = 0 \), then prove that \( a^2 + 2ac = b^2 \). 
Answer: Sum of roots \( = \sin\alpha + \cos\alpha = -\frac{b}{a} \)
Product of roots \( = \sin\alpha\cos\alpha = \frac{c}{a} \)
Squaring both sides of the sum:
\( (\sin\alpha + \cos\alpha)^2 = \left(-\frac{b}{a}\right)^2 \)
\( \implies \sin^2\alpha + \cos^2\alpha + 2\sin\alpha\cos\alpha = \frac{b^2}{a^2} \)
\( \implies 1 + 2\left(\frac{c}{a}\right) = \frac{b^2}{a^2} \)
\( \implies \frac{a + 2c}{a} = \frac{b^2}{a^2} \)
\( \implies a(a + 2c) = b^2 \)
\( \implies a^2 + 2ac = b^2 \)
Hence proved.

Question. If \( -5 \) is a root 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: Since \( -5 \) is a root of \( 2x^2 + px - 15 = 0 \):
\( 2(-5)^2 + p(-5) - 15 = 0 \)
\( \implies 50 - 5p - 15 = 0 \)
\( \implies 35 = 5p \implies p = 7 \)
Now, the equation \( p(x^2 + x) + k = 0 \) becomes \( 7x^2 + 7x + k = 0 \).
For equal roots, \( D = 0 \):
\( (7)^2 - 4(7)(k) = 0 \)
\( \implies 49 - 28k = 0 \)
\( \implies 28k = 49 \)
\( \implies k = \frac{49}{28} = \frac{7}{4} \)

Question. The coefficient of \( x \) in the quadratic equation \( x^2 + bx + c = 0 \) was taken as 17 in place of 13, its roots were found to be \( -2 \) and \( -15 \). Find the roots of the original equation.
Answer: For the incorrect equation:
Sum of roots \( = -2 + (-15) = -17 \). This matches the incorrect \( b = 17 \).
Product of roots \( = (-2)(-15) = 30 \). This gives \( c = 30 \).
The original equation is \( x^2 + 13x + 30 = 0 \).
\( \implies x^2 + 10x + 3x + 30 = 0 \)
\( \implies x(x + 10) + 3(x + 10) = 0 \)
\( \implies (x + 3)(x + 10) = 0 \)
The roots are \( -3 \) and \( -10 \).

Question. If the ratio of the roots of the equation \( lx^2 + nx + n = 0 \) is \( p : q \). Prove that \( \sqrt{\frac{p}{q}} + \sqrt{\frac{q}{p}} + \sqrt{\frac{n}{l}} = 0 \).
Answer: Let the roots be \( p\alpha \) and \( q\alpha \).
Sum of roots \( = (p + q)\alpha = -\frac{n}{l} \)
Product of roots \( = pq\alpha^2 = \frac{n}{l} \)
L.H.S. \( = \sqrt{\frac{p}{q}} + \sqrt{\frac{q}{p}} + \sqrt{\frac{n}{l}} \)
\( = \frac{p + q}{\sqrt{pq}} + \sqrt{\frac{n}{l}} \)
From the product of roots, \( \sqrt{pq}\alpha = \sqrt{\frac{n}{l}} \), so \( \sqrt{pq} = \frac{1}{\alpha}\sqrt{\frac{n}{l}} \).
Substituting this:
\( = \frac{p + q}{\frac{1}{\alpha}\sqrt{\frac{n}{l}}} + \sqrt{\frac{n}{l}} = \frac{(p + q)\alpha}{\sqrt{\frac{n}{l}}} + \sqrt{\frac{n}{l}} \)
Substituting \( (p + q)\alpha = -\frac{n}{l} \):
\( = \frac{-\frac{n}{l}}{\sqrt{\frac{n}{l}}} + \sqrt{\frac{n}{l}} = -\sqrt{\frac{n}{l}} + \sqrt{\frac{n}{l}} = 0 \)
Hence proved.

Question. If \( p, q \) are real and \( p \neq q \), then show that the roots of the equation \( (p - q)x^2 + 5(p + q)x - 2(p - q) = 0 \) are real and unequal. 
Answer: \( D = [5(p + q)]^2 - 4(p - q)[-2(p - q)] \)
\( = 25(p + q)^2 + 8(p - q)^2 \)
Since squares are always non-negative and \( p \neq q \), \( (p - q)^2 > 0 \).
Thus, \( D > 0 \).
Therefore, the roots are real and unequal.

Question. If the roots of the equation \( x^2 + 2cx + ab = 0 \) are real and unequal, prove that the equation \( x^2 - 2(a + b)x + a^2 + b^2 + 2c^2 = 0 \) has no real roots. 
Answer: For the first equation, \( D_1 = (2c)^2 - 4ab = 4c^2 - 4ab > 0 \implies c^2 - ab > 0 \).
For the second equation, \( D_2 = [-2(a + b)]^2 - 4(a^2 + b^2 + 2c^2) \)
\( = 4(a^2 + b^2 + 2ab) - 4a^2 - 4b^2 - 8c^2 \)
\( = 4a^2 + 4b^2 + 8ab - 4a^2 - 4b^2 - 8c^2 \)
\( = 8ab - 8c^2 = -8(c^2 - ab) \)
Since \( c^2 - ab > 0 \), \( D_2 < 0 \).
Hence, the second equation has no real roots.

Question. The equation \( 5x^2 + (9 + 4p)x + 2p^2 = 0 \) and \( 5x + 9 = 0 \) are satisfied by the same value of \( x \). Find the value of \( p \).
Answer: From \( 5x + 9 = 0 \), \( x = -\frac{9}{5} \).
Substituting this value into the first equation:
\( 5\left(-\frac{9}{5}\right)^2 + (9 + 4p)\left(-\frac{9}{5}\right) + 2p^2 = 0 \)
\( \implies 5\left(\frac{81}{25}\right) - \frac{9(9 + 4p)}{5} + 2p^2 = 0 \)
\( \implies \frac{81}{5} - \frac{81 + 36p}{5} + 2p^2 = 0 \)
\( \implies \frac{81 - 81 - 36p}{5} + 2p^2 = 0 \)
\( \implies -\frac{36p}{5} + 2p^2 = 0 \)
\( \implies 2p^2 - \frac{36p}{5} = 0 \implies 2p\left(p - \frac{18}{5}\right) = 0 \)
\( \implies p = 0 \) or \( p = \frac{18}{5} = 3.6 \)

Question. A tourist has Rs. 10,000 with him. He calculated that he could spend Rs. x everyday on his holidays. He spent Rs. (x - 50) everyday and extended his holidays by 10 days. Calculate x.
Answer: Original number of days \( = \frac{10000}{x} \)
New number of days \( = \frac{10000}{x - 50} \)
Difference \( = 10 \) days.
\( \frac{10000}{x - 50} - \frac{10000}{x} = 10 \)
\( \implies \frac{1000}{x - 50} - \frac{1000}{x} = 1 \)
\( \implies 1000x - 1000(x - 50) = x(x - 50) \)
\( \implies 1000x - 1000x + 50000 = x^2 - 50x \)
\( \implies x^2 - 50x - 50000 = 0 \)
\( \implies (x - 250)(x + 200) = 0 \)
Since \( x \) must be positive, \( x = 250 \).
So, \( x = 250 \).

QUESTIONS

Question. Which of the following is a solution of the equation \( x^2 - 6x + 5 = 0 \)?
(a) 2
(b) 5
(c) 9
(d) 15
Answer: (b) 5

Question. The roots of the quadratic equation \( x^2 + 5x - (\alpha + 1)(\alpha + 6) = 0 \), where \( \alpha \) is a constant, are
(a) \( \alpha + 1, \alpha + 6 \)
(b) \( (\alpha + 1), -(\alpha + 6) \)
(c) \( -(\alpha + 1), (\alpha + 6) \)
(d) \( -(\alpha + 1), -(\alpha + 6) \)
Answer: (b) \( (\alpha + 1), -(\alpha + 6) \)

Question. If a non zero root of the equations \( x^2 + 2x + 3\lambda = 0 \) and \( 2x^2 + 3x + 5\lambda = 0 \) is common, the value of \( \lambda \) will be
(a) 2
(b) 1
(c) -1
(d) 0
Answer: (c) -1

Question. Let \( \alpha, \beta \) be the roots of the equation \( (x - a)(x - b) + c = 0, c \neq 0 \). The roots of the equation \( (x - \alpha)(x - \beta) - c = 0 \) are
(a) a, c
(b) b, c
(c) a, b
(d) a + c, b + c
Answer: (c) a, b

Question. If \( \alpha, \beta \) are the roots of the equation \( x^2 - p(x + 1) - c = 0 \), then \( (\alpha + 1)(\beta + 1) = \)
(a) c
(b) c - 1
(c) 1 - c
(d) 1 + c
Answer: (c) 1 - c

Question. If \( x^2 + px + q = 0 \) is the quadratic equation whose roots are \( a - 2 \) and \( b - 2 \), where \( a, b \) are the roots of \( x^2 - 3x + 1 = 0 \), then
(a) p = 1, q = 2
(b) p = 2, q = 1
(c) p = - 1, q = 1
(d) p = 1, q = - 1
Answer: (d) p = 1, q = - 1

Question. The quadratic equation whose roots are twice the roots of \( 2x^2 - 5x + 2 = 0 \) is
(a) \( 8x^2 - 10x + 2 = 0 \)
(b) \( x^2 - 4x + 4 = 0 \)
(c) \( x^2 - 5x + 4 = 0 \)
(d) \( 2x^2 - 5x + 2 = 0 \)
Answer: (c) \( x^2 - 5x + 4 = 0 \)

Question. A quadratic equation, the product and sum of whose zeroes are 5 and 8 respectively is
(a) \( x^2 - 8x + 5 = 0 \)
(b) \( x^2 + 8x + 5 = 0 \)
(c) \( x^2 - 5x + 8 = 0 \)
(d) \( x^2 + 5x + 8 = 0 \)
Answer: (a) \( x^2 - 8x + 5 = 0 \)

Question. Which of these is a QUADRATIC equation having one of its roots as zero?
(i) \( x^3 + x^2 = 0 \)
(ii) \( x^2 - 2x = 0 \)
(iii) \( x^2 - 9 = 0 \)
(a) only (i)
(b) only (ii)
(c) only (i) and (ii)
(d) only (ii) and (iii)
Answer: (b) only (ii)

Question. Three students were asked how they would verify their solution of a quadratic equation, \( (x - 2)(x - 5) = 0 \). Shown below are their responses.
Student 1 said, "In the first bracket, \( x \) must equal 2, and in the second bracket, \( x \) must equal 5. So, \( (2 - 2)(5 - 5) = 0 \)."
Student 2 said, "In the first bracket, \( x \) must equal 2, but in the second bracket, \( x \) can have any real number value. For example, \( (2 - 2)(3 - 5) = 0 \) or \( (2 - 2)(10 - 5) = 0 \)."
Student 3 said, "Both brackets should always have the same \( x \) value. So, \( x \) is either 2 or 5 in both brackets. For example, \( (2 - 2)(2 - 5) = 0 \) and \( (5 - 2)(5 - 5) = 0 \)."
Whose response is correct?
(a) only student 1
(b) only student 3
(c) only students 1 and 2
(d) all students 1, 2 and 3
Answer: (b) only student 3

Question. If \( x = k\sqrt{2} \) be a solution of the quadratic \( x^2 + \sqrt{2}x - 4 = 0 \), then \( k = \)
(a) -1
(b) -2
(c) 2
(d) 4
Answer: (b) -2

Question. The non-negative real root of the quadratic equation \( 3x^2 - 5x - 2 = 0 \) is
(a) 3
(b) \( \frac{1}{3} \)
(c) 2
(d) \( \frac{1}{2} \)
Answer: (c) 2

Question. If arithmetic mean of two numbers \( a \) and \( b \) is \( A \) and \( a \cdot b = G^2 \) then a quadratic equation whose roots are \( a \) and \( b \) is
(a) \( x^2 - 2Ax + G^2 = 0 \)
(b) \( x^2 + 2Ax + G^2 = 0 \)
(c) \( x^2 - 2Ax - G^2 = 0 \)
(d) \( x^2 + 2Ax - G^2 = 0 \)
Answer: (a) \( x^2 - 2Ax + G^2 = 0 \)

Question. The discriminant of the quadratic equation \( ax^2 - 4ax + 2a + 1 = 0 \) is equal to
(a) \( 4a(2a + 1) \)
(b) \( 2a(2a + 1) \)
(c) \( 4a(2a - 1) \)
(d) \( 2a(4a - 1) \)
Answer: (c) \( 4a(2a - 1) \)

Question. The quadratic equation \( ax^2 - 4ax + 2a + 1 = 0 \) has repeated root if \( a = \)
(a) 0
(b) \( \frac{1}{2} \)
(c) 2
(d) 4
Answer: (b) \( \frac{1}{2} \)

Question. If \( x^2 - 5x + 1 = 0 \), then the value of \( x + \frac{1}{x} \) is
(a) 5
(b) -5
(c) -2
(d) 3
Answer: (a) 5

Question. The quadratic \( kx^2 - 2kx + 2 = 0 \) has equal roots if \( k = \)
(a) 0
(b) 2
(c) 1
(d) 4
Answer: (b) 2

Question. Which of the following equations has the sum of its roots as 3?
(a) \( 2x^2 - 3x + 6 = 0 \)
(b) \( -x^2 + 3x - 3 = 0 \)
(c) \( \sqrt{2}x^2 - \frac{3}{\sqrt{2}}x + 1 = 0 \)
(d) \( 3x^2 - 3x + 3 = 0 \)
Answer: (b) \( -x^2 + 3x - 3 = 0 \)

Question. If \( px^2 + 3x + q = 0 \) has two roots -1 and -2, then the value of \( q - p \) is
(a) -1
(b) 2
(c) -2
(d) 1
Answer: (d) 1

Question. The quadratic equation with integral coefficients, whose one root is \( 2 + \sqrt{3} \), is
(a) \( x^2 - 4x + 1 = 0 \)
(b) \( x^2 - 2x + 1 = 0 \)
(c) \( x^2 - 4x + 4 = 0 \)
(d) \( x^2 - 4x + 3 = 0 \)
Answer: (a) \( x^2 - 4x + 1 = 0 \)

Question. The common root of the quadratic equations \( x^2 - 3x + 2 = 0 \) and \( 2x^2 - 5x + 2 = 0 \) is
(a) 2
(b) 1
(c) -2
(d) \( \frac{1}{2} \)
Answer: (a) 2

Question. Which of the following equations has two distinct real roots?
(a) \( 2x^2 - 3\sqrt{2}x + \frac{9}{4} = 0 \)
(b) \( x^2 + x - 5 = 0 \)
(c) \( x^2 + 3x + 2\sqrt{2} = 0 \)
(d) \( 5x^2 - 3x + 1 = 0 \)
Answer: (b) \( x^2 + x - 5 = 0 \)

Question. The equation \( (x^2 + 1)^2 - x^2 = 0 \) has
(a) four real roots
(b) two real roots
(c) no real roots
(d) one real roots
Answer: (c) no real roots

Question. If \( x > y > 0 \), \( x^2 + y^2 = 13 \) and \( xy = 6 \), then \( y = \)
(a) 4
(b) 3
(c) 2
(d) 1
Answer: (c) 2

Question. \( S_1 \) and \( S_2 \) are two square figures. \( S_1 \) is smaller square and length of its side is \( x \) cm; \( S_2 \) is bigger square and length of its side is \( y \) cm. If it is given that \( x^2 + y^2 = 74 \) and \( 2x^2 - y^2 = 1 \) then the side of the larger square is of length
(a) 5 cm
(b) 7 cm
(c) 25 cm
(d) 14 cm
Answer: (b) 7 cm

Question. If the roots of the quadratic equation \( kx^2 + (a + b)x + ab = 0 \) are -a and -b, then the value of \( k \) is
(a) -1
(b) 1
(c) 2
(d) -2
Answer: (b) 1

Question. Solve the quadratic equation \( 2x^2 + ax - a^2 = 0 \) for \( x \). 
Answer: \( 2x^2 + ax - a^2 = 0 \)
\( \implies 2x^2 + 2ax - ax - a^2 = 0 \)
\( \implies 2x(x + a) - a(x + a) = 0 \)
\( \implies (2x - a)(x + a) = 0 \)
\( \implies x = \frac{a}{2}, -a \)