CBSE Class 10 Maths HOTs Quadratic Equations Set 18

Question. Solve the following quadratic equation for \( x \): \( 4x^2 - 4a^2x + (a^4 - b^4) = 0 \)
Answer: \( 4x^2 - 4a^2x + (a^2 - b^2)(a^2 + b^2) = 0 \)
\( \implies 4x^2 - [2(a^2 - b^2) + 2(a^2 + b^2)]x + (a^2 - b^2)(a^2 + b^2) = 0 \)
\( \implies 4x^2 - 2(a^2 - b^2)x - 2(a^2 + b^2)x + (a^2 - b^2)(a^2 + b^2) = 0 \)
\( \implies 2x[2x - (a^2 - b^2)] - (a^2 + b^2)[2x - (a^2 - b^2)] = 0 \)
\( \implies [2x - (a^2 - b^2)][2x - (a^2 + b^2)] = 0 \)
\( \implies x = \frac{a^2 - b^2}{2}, \frac{a^2 + b^2}{2} \)

Question. Solve for \( x \): \( x^2 - (\sqrt{3} + 1)x + \sqrt{3} = 0 \) 
Answer: \( x^2 - \sqrt{3}x - x + \sqrt{3} = 0 \)
\( \implies x(x - \sqrt{3}) - 1(x - \sqrt{3}) = 0 \)
\( \implies (x - 1)(x - \sqrt{3}) = 0 \)
\( \implies x = 1, \sqrt{3} \)

Question. Solve for \( x \): \( \sqrt{2x + 9} + x = 13 \). 
Answer: \( \sqrt{2x + 9} = 13 - x \)
Squaring both sides,
\( 2x + 9 = (13 - x)^2 \)
\( \implies 2x + 9 = 169 + x^2 - 26x \)
\( \implies x^2 - 28x + 160 = 0 \)
\( \implies x^2 - 20x - 8x + 160 = 0 \)
\( \implies x(x - 20) - 8(x - 20) = 0 \)
\( \implies (x - 8)(x - 20) = 0 \)
\( \implies x = 8, 20 \)
Check \( x = 20 \): \( \sqrt{2(20) + 9} + 20 = \sqrt{49} + 20 = 7 + 20 = 27 \neq 13 \). (Rejected)
Check \( x = 8 \): \( \sqrt{2(8) + 9} + 8 = \sqrt{25} + 8 = 5 + 8 = 13 \).
\( \therefore x = 8 \)

Question. A two-digit number is four times the sum of its digits. It is also equal to 3 times the product of digits. Find the number. 
Answer: Let the tens digit be \( x \) and units digit be \( y \).
Number \( = 10x + y \)
\( 10x + y = 4(x + y) \implies 6x = 3y \implies y = 2x \)
Also, \( 10x + y = 3xy \)
\( \implies 10x + 2x = 3x(2x) \)
\( \implies 12x = 6x^2 \implies 2 = x \) (since \( x \neq 0 \))
If \( x = 2 \), then \( y = 4 \).
The number is 24.

Question. Solve for \( x \): \( (a + b)^2x^2 + 8(a^2 - b^2)x + 16(a - b)^2 = 0 \)
Answer: This is in the form \( (Ax + B)^2 = 0 \).
\( [(a + b)x + 4(a - b)]^2 = (a + b)^2x^2 + 16(a - b)^2 + 8(a + b)(a - b)x = 0 \)
\( \implies (a + b)x + 4(a - b) = 0 \)
\( \implies x = \frac{-4(a - b)}{a + b} = \frac{4(b - a)}{a + b} \)

Question. Find the value of \( k \) for which the quadratic equation \( (k + 4)x^2 + (k + 1)x + 1 = 0 \) has equal roots.
Answer: For equal roots, \( D = 0 \).
\( (k + 1)^2 - 4(k + 4)(1) = 0 \)
\( \implies k^2 + 2k + 1 - 4k - 16 = 0 \)
\( \implies k^2 - 2k - 15 = 0 \)
\( \implies (k - 5)(k + 3) = 0 \)
\( \implies k = 5, -3 \)

Question. Solve for \( x \): \( x^2 - (\sqrt{2} + 1)x + \sqrt{2} = 0 \)
Answer: \( x^2 - \sqrt{2}x - x + \sqrt{2} = 0 \)
\( \implies x(x - \sqrt{2}) - 1(x - \sqrt{2}) = 0 \)
\( \implies (x - 1)(x - \sqrt{2}) = 0 \)
\( \implies x = 1, \sqrt{2} \)

Question. Using quadratic formula, solve the following quadratic equation for \( x \): \( x^2 - 4ax + 4a^2 - b^2 = 0 \).
Answer: \( a = 1, b = -4a, c = 4a^2 - b^2 \)
\( D = (-4a)^2 - 4(1)(4a^2 - b^2) = 16a^2 - 16a^2 + 4b^2 = 4b^2 \)
\( x = \frac{-(-4a) \pm \sqrt{4b^2}}{2(1)} = \frac{4a \pm 2b}{2} = 2a \pm b \)
\( \implies x = 2a + b, 2a - b \)

Question. Solve the quadratic equation: \( x^2 + 2\sqrt{2}x - 6 = 0 \) for \( x \). 
Answer: \( x^2 + 3\sqrt{2}x - \sqrt{2}x - 6 = 0 \)
\( \implies x(x + 3\sqrt{2}) - \sqrt{2}(x + 3\sqrt{2}) = 0 \)
\( \implies (x - \sqrt{2})(x + 3\sqrt{2}) = 0 \)
\( \implies x = \sqrt{2}, -3\sqrt{2} \)

Question. If one root of the quadratic equation \( 2x^2 - 3x + p = 0 \) is 3, find the other root of the quadratic equation. Also find the value of \( p \).
Answer: Putting \( x = 3 \) in \( 2x^2 - 3x + p = 0 \):
\( 2(3)^2 - 3(3) + p = 0 \implies 18 - 9 + p = 0 \implies p = -9 \).
Sum of roots \( = \alpha + \beta = -(-3)/2 = 3/2 \).
Given \( \alpha = 3 \), so \( 3 + \beta = 3/2 \implies \beta = 3/2 - 3 = -3/2 \).
The other root is -3/2 and \( p = -9 \).

Question. One root of the equation \( 2x^2 - 8x - m = 0 \) is 5/2. Find the other root and the value of \( m \).
Answer: Putting \( x = 5/2 \):
\( 2(5/2)^2 - 8(5/2) - m = 0 \implies 25/2 - 20 - m = 0 \implies m = 12.5 - 20 = -7.5 \).
Sum of roots \( = 8/2 = 4 \).
Let other root be \( \beta \). \( 5/2 + \beta = 4 \implies \beta = 4 - 2.5 = 1.5 = 3/2 \).
Other root is 3/2 and \( m = -7.5 \) (or \( -15/2 \)).

Question. The sum of a number and its positive square root is 6/25. Find the number.
Answer: Let the number be \( x \). \( x + \sqrt{x} = 6/25 \).
Let \( \sqrt{x} = y \). Then \( y^2 + y - 6/25 = 0 \implies 25y^2 + 25y - 6 = 0 \).
\( 25y^2 + 30y - 5y - 6 = 0 \implies 5y(5y + 6) - 1(5y + 6) = 0 \).
\( \implies y = 1/5, -6/5 \). Since \( y = \sqrt{x} \), \( y > 0 \), so \( y = 1/5 \).
\( \therefore x = (1/5)^2 = 1/25 \).

Question. For what values of \( p \) the equation \( (1 + p)x^2 + 2(1 + 2p)x + (1 + p) = 0 \) has coincident roots?
Answer: For coincident roots, \( D = 0 \).
\( [2(1 + 2p)]^2 - 4(1 + p)(1 + p) = 0 \)
\( \implies 4(1 + 4p^2 + 4p) - 4(1 + p)^2 = 0 \)
\( \implies 1 + 4p^2 + 4p - (1 + 2p + p^2) = 0 \)
\( \implies 3p^2 + 2p = 0 \implies p(3p + 2) = 0 \)
\( \implies p = 0 \) or \( p = -2/3 \).

Question. In each of the following determine whether the given values are solutions of the equations or not:
(i) \( 6x^2 - x - 2 = 0 \); \( x = -1/2 \); \( x = 2/3 \)
(ii) \( x^2 + \sqrt{2}x - 4 = 0 \), \( x = \sqrt{2} \), \( x = -2\sqrt{2} \)
Answer: (i) Check \( x = -1/2 \): \( 6(1/4) - (-1/2) - 2 = 3/2 + 1/2 - 2 = 0 \) (Solution).
Check \( x = 2/3 \): \( 6(4/9) - 2/3 - 2 = 8/3 - 2/3 - 2 = 6/3 - 2 = 0 \) (Solution).
(ii) Check \( x = \sqrt{2} \): \( (\sqrt{2})^2 + \sqrt{2}(\sqrt{2}) - 4 = 2 + 2 - 4 = 0 \) (Solution).
Check \( x = -2\sqrt{2} \): \( (-2\sqrt{2})^2 + \sqrt{2}(-2\sqrt{2}) - 4 = 8 - 4 - 4 = 0 \) (Solution).

Question. Solve the equation \( \frac{4}{x} - 3 = \frac{5}{2x + 3} \); \( x \neq 0, -\frac{3}{2} \), for \( x \). 
Answer: \( \frac{4 - 3x}{x} = \frac{5}{2x + 3} \)
\( \implies (4 - 3x)(2x + 3) = 5x \)
\( \implies 8x + 12 - 6x^2 - 9x = 5x \)
\( \implies -6x^2 - x + 12 = 5x \)
\( \implies 6x^2 + 6x - 12 = 0 \implies x^2 + x - 2 = 0 \)
\( \implies (x + 2)(x - 1) = 0 \implies x = 1, -2 \)

Question. Solve for \( x \): \( x^2 + 5x - (a^2 + a - 6) = 0 \)
Answer: \( a^2 + a - 6 = (a + 3)(a - 2) \).
Equation: \( x^2 + [(a + 3) - (a - 2)]x - (a + 3)(a - 2) = 0 \)
\( \implies x^2 + (a + 3)x - (a - 2)x - (a + 3)(a - 2) = 0 \)
\( \implies x[x + (a + 3)] - (a - 2)[x + (a + 3)] = 0 \)
\( \implies [x - (a - 2)][x + (a + 3)] = 0 \)
\( \implies x = a - 2, -(a + 3) \)

Question. One root of the quadratic equation \( 2x^2 - 8x - k = 0 \) is 5/2. Find the value of \( k \). Also, find the other root.
Answer: Putting \( x = 5/2 \):
\( 2(5/2)^2 - 8(5/2) - k = 0 \implies 25/2 - 20 - k = 0 \implies k = 12.5 - 20 = -7.5 \).
Sum of roots \( = 8/2 = 4 \).
Let other root be \( \beta \). \( 5/2 + \beta = 4 \implies \beta = 4 - 2.5 = 1.5 = 3/2 \).
Other root is 3/2 and \( k = -7.5 \).

Question. Using quadratic formula, solve the following equation for \( x \): \( abx^2 + (b^2 - ac) x - bc = 0 \)
Answer: \( D = (b^2 - ac)^2 - 4(ab)(-bc) = b^4 + a^2c^2 - 2ab^2c + 4ab^2c = b^4 + a^2c^2 + 2ab^2c = (b^2 + ac)^2 \)
\( x = \frac{-(b^2 - ac) \pm \sqrt{(b^2 + ac)^2}}{2ab} = \frac{-b^2 + ac \pm (b^2 + ac)}{2ab} \)
\( \implies x = \frac{-b^2 + ac + b^2 + ac}{2ab} = \frac{2ac}{2ab} = \frac{c}{b} \)
and \( x = \frac{-b^2 + ac - b^2 - ac}{2ab} = \frac{-2b^2}{2ab} = -\frac{b}{a} \)

Question. A two-digit number is such that the product of its digits is 15. If 18 is added to the number, the digits interchange their places. Find the number.
Answer: Let digits be \( x \) and \( y \). Number \( = 10x + y \).
\( xy = 15 \).
\( 10x + y + 18 = 10y + x \implies 9y - 9x = 18 \implies y - x = 2 \implies y = x + 2 \).
\( x(x + 2) = 15 \implies x^2 + 2x - 15 = 0 \implies (x + 5)(x - 3) = 0 \implies x = 3 \) (as \( x > 0 \)).
If \( x = 3 \), then \( y = 5 \).
The number is 35.

Question. The area of a right-angled triangle is 600 sq cm. If the base of the triangle exceeds the altitude by 10 cm, find the dimensions of the triangle.
Answer: Let altitude be \( h \). Then base \( = h + 10 \).
\( \frac{1}{2} \cdot h(h + 10) = 600 \implies h^2 + 10h - 1200 = 0 \).
\( \implies h^2 + 40h - 30h - 1200 = 0 \implies (h + 40)(h - 30) = 0 \implies h = 30 \).
Altitude \( = 30 \) cm, base \( = 40 \) cm.
Hypotenuse \( = \sqrt{30^2 + 40^2} = 50 \) cm.

Question. The cost price of an article is Rs. x and is sold at a profit of \( (x + 10) \) %. Find the cost price of the article, if its selling price is Rs. \( (2x - 20) \).
Answer: \( S.P. = C.P. \times (1 + Profit\%) \)
\( 2x - 20 = x \cdot \left(1 + \frac{x + 10}{100}\right) \)
\( \implies 2x - 20 = x \cdot \left(\frac{110 + x}{100}\right) \)
\( \implies 200x - 2000 = 110x + x^2 \)
\( \implies x^2 - 90x + 2000 = 0 \implies (x - 40)(x - 50) = 0 \)
\( \implies x = 40 \) or \( x = 50 \).
The cost price is Rs. 40 or Rs. 50.

Question. The product of two successive integral multiples of 5 is 300. Determine the multiples.
Answer: Let multiples be \( 5n \) and \( 5(n + 1) \).
\( 5n \cdot 5(n + 1) = 300 \implies 25(n^2 + n) = 300 \implies n^2 + n - 12 = 0 \).
\( \implies (n + 4)(n - 3) = 0 \implies n = 3, -4 \).
If \( n = 3 \), multiples are 15, 20.
If \( n = -4 \), multiples are -20, -15.

Question. Solve the equation: \( \frac{4x}{x - 2} - \frac{3x}{x - 1} = 7 \frac{1}{2} \).
Answer: \( \frac{4x(x - 1) - 3x(x - 2)}{(x - 2)(x - 1)} = \frac{15}{2} \)
\( \implies \frac{4x^2 - 4x - 3x^2 + 6x}{x^2 - 3x + 2} = \frac{15}{2} \)
\( \implies 2(x^2 + 2x) = 15(x^2 - 3x + 2) \)
\( \implies 2x^2 + 4x = 15x^2 - 45x + 30 \)
\( \implies 13x^2 - 49x + 30 = 0 \)
\( \implies 13x^2 - 39x - 10x + 30 = 0 \)
\( \implies 13x(x - 3) - 10(x - 3) = 0 \implies x = 3, 10/13 \).

Question. Solve the equation: \( 2(x - 3)^2 + 3(x - 2)(2x - 3) = 8(x + 4)(x - 4) - 1 \).
Answer: \( 2(x^2 - 6x + 9) + 3(2x^2 - 7x + 6) = 8(x^2 - 16) - 1 \)
\( \implies 2x^2 - 12x + 18 + 6x^2 - 21x + 18 = 8x^2 - 128 - 1 \)
\( \implies 8x^2 - 33x + 36 = 8x^2 - 129 \)
\( \implies -33x = -129 - 36 \implies -33x = -165 \implies x = 5 \).

Question. The age of father is equal to the square of the age of his son. The sum of the age of father and five times the age of the son is 66 years. Find their ages.
Answer: Let son's age be \( x \). Father's age \( = x^2 \).
\( x^2 + 5x = 66 \implies x^2 + 5x - 66 = 0 \).
\( \implies (x + 11)(x - 6) = 0 \implies x = 6 \) (as age cannot be negative).
Son's age \( = 6 \) years, Father's age \( = 36 \) years.

ASSERTION AND REASON QUESTIONS

In the following questions, a statement of assertion (A) is followed by a statement of Reason (R). Choose the correct answer out of the following choices.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question. Assertion (A): \( 3y^2 + 17y - 30 = 0 \) have distinct roots
Reason (R): The quadratic equation \( ax^2 + bx + c = 0 \) have distinct roots (real roots) if \( D > 0 \).
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (a) Both A and R are true and R is the correct explanation of A.

Question. Assertion (A): \( 2\sqrt{2} \) is not the root of the quadratic equation \( x^2 - 4\sqrt{2}x + 8 = 0 \).
Reason (R): The root of a quadratic satisfies it.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (d) A is false but R is true.