CBSE Class 10 Mathematics Quadratic Equations Worksheet Set 07

Question. Find the roots of the quadratic equation \( \sqrt{2}x^2 + 7x + 5\sqrt{2} = 0 \).
Answer: The given quadratic equation is \( \sqrt{2}x^2 + 7x + 5\sqrt{2} = 0 \)
By applying mid term splitting, we get
\( \sqrt{2}x^2 + 2x + 5x + 5\sqrt{2} = 0 \)

\( \implies \) \( \sqrt{2}x(x + \sqrt{2}) + 5(x + \sqrt{2}) = 0 \)

\( \implies \) \( (\sqrt{2}x + 5)(x + \sqrt{2}) = 0 \)

\( \implies \) \( x = \frac{-5}{\sqrt{2}}, -\sqrt{2} \) or \( \frac{-5\sqrt{2}}{2}, -\sqrt{2} \)

Question. A train covers a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. Find the original speed of the train. 
Answer: Let the speed of train be \( x \) km/h.
\(\therefore\) \( \frac{480}{x - 8} - \frac{480}{x} = 3 \)

\( \implies \) \( x^2 - 8x - 1280 = 0 \)
\( (x - 40)(x + 32) = 0 \)
\( x = 40, -32 \text{ (Rejected)} \)
\(\therefore\) Speed of train = 40 km/h

Question. In a class test, the sum of Shefali's marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of her marks would have been 210. Find her marks in the two subjects.
Answer: Let Shefali's marks in Mathematics be \( x \).
Therefore, Shefali's marks in English is \( (30 - x) \).
Now, according to question,
\( (x + 2) (30 - x - 3) = 210 \)

\( \implies \) \( (x + 2) (27 - x) = 210 \)

\( \implies \) \( 27x - x^2 + 54 - 2x = 210 \)

\( \implies \) \( 25x - x^2 - 156 = 0 \)

\( \implies \) \( x^2 - 25x + 156 = 0 \)

\( \implies \) \( x^2 - 13x - 12x + 156 = 0 \)

\( \implies \) \( (x - 13) (x - 12) = 0 \)
Either \( x - 13 = 0 \) or \( x - 12 = 0 \)

\( \implies \) \( x = 13 \) or \( x = 12 \)
\(\therefore\) Shefali's marks in Mathematics = 13, marks in English = \( 30 - 13 = 17 \)
or Shefali's marks in Mathematics = 12, marks in English = \( 30 - 12 = 18 \).

Question. Find the roots of the following quadratic equation by applying the quadratic formula.
\( 4x^2 + 4\sqrt{3}x + 3 = 0 \)
Answer: We have, \( 4x^2 + 4\sqrt{3}x + 3 = 0 \)
Here, \( a = 4, b = 4\sqrt{3} \) and \( c = 3 \)
Therefore, \( D = b^2 - 4ac \)
\( = (4\sqrt{3})^2 - 4 \times 4 \times 3 = 48 - 48 = 0 \)
\(\because\) \( D = 0 \), roots exist and are equal.
Thus, \( x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-4\sqrt{3} \pm 0}{2 \times 4} = \frac{-\sqrt{3}}{2} \)
Hence, the roots of given equation are \( \frac{-\sqrt{3}}{2} \) and \( \frac{-\sqrt{3}}{2} \).

Question. Find the roots of the following quadratic equation by factorisation:
\( 2x^2 - x + \frac{1}{8} = 0 \)
Answer: We have, \( 2x^2 - x + \frac{1}{8} = 0 \)

\( \implies \) \( \frac{16x^2 - 8x + 1}{8} = 0 \)

\( \implies \) \( 16x^2 - 8x + 1 = 0 \)

\( \implies \) \( 16x^2 - 4x - 4x + 1 = 0 \)

\( \implies \) \( 4x (4x - 1) - 1 (4x - 1) = 0 \)

\( \implies \) \( (4x - 1) (4x - 1) = 0 \)
So, either \( 4x - 1 = 0 \) or \( 4x - 1 = 0 \)
\( x = \frac{1}{4} \) or \( x = \frac{1}{4} \)
Hence, the roots of the given equation are \( \frac{1}{4} \) and \( \frac{1}{4} \).

Question. A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream. 
Answer: Given that:
Speed of boat = 18 km/hr in still water.
Speed of stream = \( S \) (variable, must find)
Distance upstream = 24 km
Time upstream = 1 hr more than time downstream.
We know, \( \text{speed} = \frac{\text{Distance}}{\text{Time}} \implies \text{Time} = \frac{\text{Distance}}{\text{speed}} \).
Time upstream = \( \frac{24}{18 - S} \), Time downstream = \( \frac{24}{18 + S} \).

\( \implies \) \( \frac{24}{18 - S} = 1 + \frac{24}{18 + S} \)

\( \implies \) \( \frac{24}{18 - S} = \frac{18 + S + 24}{18 + S} \)

\( \implies \) \( 24(18 + S) = (42 + S)(18 - S) \) (Cross-multiplying)

\( \implies \) \( 432 + 24S = 756 + 18S - 42S - S^2 \)

\( \implies \) \( S^2 + 24S + 24S + 432 - 756 = 0 \)

\( \implies \) \( S^2 + 48S - 324 = 0 \)

\( \implies \) \( S^2 + 54S - 6S - 324 = 0 \)

\( \implies \) \( S(S + 54) - 6(S + 54) = 0 \)

\( \implies \) \( (S - 6)(S + 54) = 0 \)
Now, either \( S - 6 = 0 \) or \( S + 54 = 0 \).

\( \implies \) \( S = 6 \) or \( S = -54 \).
So speed = 6 or -54 km/hr. But speed cannot be negative.

\( \implies \) Speed of the stream is 6 km/hr.
[Topper's Answer 2018(30/1/1)]

Question. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
Answer: Let \( ABCD \) be the rectangular field. Let the shorter side \( BC \) of the rectangle = \( x \) metres.
According to question,
Diagonal of the rectangle, \( AC = (x + 60) \) metres
side of the rectangle, \( AB = (x + 30) \) metres
By Pythagoras theorem, \( AC^2 = AB^2 + BC^2 \)
\(\therefore\) \( (x + 60)^2 = (x + 30)^2 + x^2 \)
or \( x^2 + 120x + 3600 = x^2 + 60x + 900 + x^2 \)
or \( 2x^2 - x^2 + 60x - 120x + 900 - 3600 = 0 \)
or \( x^2 - 60x - 2700 = 0 \)
or \( (x - 90)(x + 30) = 0 \)
Either \( x - 90 = 0 \) or \( x + 30 = 0 \)

\( \implies \) \( x = 90 \) or \( x = -30 \) (But side cannot be negative)
So, the shorter side of rectangle = 90 m and longer side of rectangle = 120 m

Question. The sum of the reciprocals of Rehman's age (in years) 3 years ago and 5 years from now is \( \frac{1}{3} \). Find his present age.
Answer: Let the present age of Rehman be \( x \) years.
So, 3 years ago, Rehman's age = \( (x - 3) \) years
And 5 years from now, Rehman's age = \( (x + 5) \) years
Now, according to question, we have
\( \frac{1}{x - 3} + \frac{1}{x + 5} = \frac{1}{3} \)
\( \implies \) \( \frac{x + 5 + x - 3}{(x - 3)(x + 5)} = \frac{1}{3} \)

\( \implies \) \( \frac{2x + 2}{(x - 3)(x + 5)} = \frac{1}{3} \)

\( \implies \) \( 6x + 6 = (x - 3)(x + 5) \)

\( \implies \) \( 6x + 6 = x^2 + 5x - 3x - 15 \)

\( \implies \) \( x^2 - 4x - 21 = 0 \)

\( \implies \) \( x(x - 7) + 3(x - 7) = 0 \)

\( \implies \) \( (x - 7)(x + 3) = 0 \)

\( \implies \) \( x = 7 \) or \( x = -3 \)
But \( x \neq -3 \) (age cannot be negative)
Therefore, present age of Rehman = 7 years.

Question. Find the value of \( k \) for which the quadratic equation \( kx(x - 2) + 6 = 0 \) has two equal roots. 
Answer: Given equation \( kx^2 - 2kx + 6 = 0 \)
For two equal roots
\( D = 0 \)

\( \implies \) \( b^2 - 4ac = 0 \)

\( \implies \) \( 4k^2 - 4k \times 6 = 0 \)

\( \implies \) \( 4k(k - 6) = 0 \)

\( \implies \) \( k = 6 \) [\( k \neq 0 \), as if \( k = 0 \) then the given equation is not a valid equation.]

Question. The difference of square of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
Answer: Let smaller and larger number be \( x \) and \( y \) respectively.
From question \( y^2 - x^2 = 180 \) ...(i)
Also, \( x^2 = 8y \) ...(ii)
From (i) and (ii)
\( y^2 - 8y - 180 = 0 \)
By applying quadratic formula, we get
\( y = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \times 1 \times (-180)}}{2} \)

\( \implies \) \( y = \frac{8 \pm \sqrt{64 + 720}}{2} = \frac{8 \pm \sqrt{784}}{2} \)

\( \implies \) \( y = \frac{8 \pm 28}{2} = 18 \text{ or } -10 \)
If \( y = 18 \), \( x^2 = 8 \times 18 = 144 \implies x = \pm 12 \)
Required two numbers are 18, 12 or 18, -12.
Also, \( y = -10 \) is not possible because \( x^2 = 8 \times (-10) = -80 \) (negative number which is not possible)

Multiple Choice Questions

Choose and write the correct option in the following questions.

Question. The product of three consecutive integers is equal to 6 times the sum of three integers. If the smallest integer is \( x \), which of the following equations represent the above situation?
(a) \( 2x^2 + x - 9 = 0 \)
(b) \( 2x^2 - x + 9 = 0 \)
(c) \( x^2 + 2x + 18 = 0 \)
(d) \( x^2 + 2x - 18 = 0 \)
Answer: (d) \( x^2 + 2x - 18 = 0 \)

Question. Consider the equation \( kx^2 + 2x = c(2x^2 + b) \). For the equation to be quadratic, which of these cannot be the value of \( k \)? 
(a) \( 2c \)
(b) \( 3c \)
(c) \( 4c \)
(d) \( 2c + 2b \)
Answer: (a) \( 2c \)

Question. Which of the following is not a quadratic equation? 
(a) \( 2(x - 1)^2 = 4x^2 - 2x + 1 \)
(b) \( 2x - x^2 = x^2 + 5 \)
(c) \( (\sqrt{2}x + \sqrt{3})^2 + x^2 = 3x^2 - 5x \)
(d) \( (x^2 + 2x)^2 = x^4 + 3 + 4x^3 \)
Answer: (c) \( (\sqrt{2}x + \sqrt{3})^2 + x^2 = 3x^2 - 5x \)

Question. The roots of the quadratic equation \( x^2 - 0.04 = 0 \) are
(a) \( \pm 0.2 \)
(b) \( \pm 0.02 \)
(c) 0.4
(d) 2
Answer: (a) \( \pm 0.2 \)

Question. Which of the following equations has 2 as a root? 
(a) \( x^2 - 4x + 5 = 0 \)
(b) \( x^2 + 3x - 12 = 0 \)
(c) \( 2x^2 - 7x + 6 = 0 \)
(d) \( 3x^2 - 6x - 2 = 0 \)
Answer: (c) \( 2x^2 - 7x + 6 = 0 \)

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. Rahul follows the below steps to find the roots of the equation \( 3x^2 - 11x - 20 = 0 \), by splitting the middle term.
Step 1: \( 3x^2 - 11x - 20 = 0 \)
Step 2: \( 3x^2 - 15x + 4x - 20 = 0 \)
Step 3: \( 3x(x - 5) + 4(x - 5) = 0 \)
Step 4: \( (3x - 4)(x - 5) = 0 \)
Step 5: \( x = \frac{4}{3} \) and 5
In which step did Rahul make the first error? 
(a) Step 1
(b) Step 2
(c) Step 3
(d) Step 4
Answer: (d) Step 4

Question. If the list price of a toy is reduced by Rs. 2, a person can buy 2 toys more for Rs. 360. The original price of the toy is
(a) Rs. 18
(b) Rs. 20
(c) Rs. 19
(d) Rs. 21
Answer: (b) Rs. 20

Question. The quadratic equation whose one rational root is \( 3 + \sqrt{2} \) is
(a) \( x^2 - 7x + 5 = 0 \)
(b) \( x^2 + 7x + 6 = 0 \)
(c) \( x^2 - 7x + 6 = 0 \)
(d) \( x^2 - 7x + 7 = 0 \)
Answer: (d) \( x^2 - 7x + 7 = 0 \)

Question. A student is trying to find the roots of \( 3x^2 - 10x - 8 = 0 \) by splitting the middle term as follows:
Step 1: \( 3x^2 - 10x - 8 = 0 \)
Step 2: \( 3x^2 - mx + nx - 8 = 0 \)
What could be the values of \( m \) and \( n \)? 
(a) \( m = 8 \) and \( n = 2 \)
(b) \( m = -8 \) and \( n = -2 \)
(c) \( m = 12 \) and \( n = 2 \)
(d) \( m = -12 \) and \( n = -2 \)
Answer: (c) \( m = 12 \) and \( n = 2 \)

Question. The value(s) of \( k \) for which the quadratic equation \( 2x^2 + kx + 2 = 0 \) has equal roots, is
(a) 4
(b) \( \pm 4 \)
(c) -4
(d) 0
Answer: (b) \( \pm 4 \)

Question. If the difference of roots of the quadratic equation \( x^2 + kx + 12 = 0 \) is 1, the positive value of \( k \) is
(a) -7
(b) 7
(c) 4
(d) 8
Answer: (b) 7

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 quadratic equation \( 2x^2 - \sqrt{5}x + 1 = 0 \) has
(a) two distinct real roots
(b) two equal real roots
(c) no real root
(d) more than two real roots
Answer: (c) no real root

Question. Which of the following equations has no real roots? 
(a) \( x^2 - 4x + 3\sqrt{2} = 0 \)
(b) \( x^2 + 4x - 3\sqrt{2} = 0 \)
(c) \( x^2 - 4x - 3\sqrt{2} = 0 \)
(d) \( 3x^2 + 4\sqrt{3}x + 4 = 0 \)
Answer: (a) \( x^2 - 4x + 3\sqrt{2} = 0 \)

Question. The quadratic equation \( x^2 - 4x + k = 0 \) has distinct real roots if 
(a) \( k = 4 \)
(b) \( k > 4 \)
(c) \( k = 16 \)
(d) \( k < 4 \)
Answer: (d) \( k < 4 \)

Question. Consider the equation \( px^2 + qx + r = 0 \). Which conditions are sufficient to conclude that the equation has real roots?
(a) \( p < 0, q > 0 \)
(b) \( p > 0, q < 0 \)
(c) \( p > 0, r > 0 \)
(d) \( p > 0, r < 0 \)
Answer: (d) \( p > 0, r < 0 \)

Question. If the equation \( x^2 - mx + 1 = 0 \) does not possess real roots, then
(a) \( m > 2 \)
(b) \( m < -2 \)
(c) \( -2 < m < 2 \)
(d) \( -3 < m < 3 \)
Answer: (c) \( -2 < m < 2 \)

Very Short Answer Questions

Each of the following questions are of 1 mark.

Question. Find the value of \( k \) for which \( x = 2 \) is a solution of the equation \( kx^2 + 2x - 3 = 0 \). 
Answer: \( k(2)^2 + 2(2) - 3 = 0 \)

\( \implies \) \( 4k + 1 = 0 \)

\( \implies \) \( k = -\frac{1}{4} \)

Question. If \( x = 3 \) is one root of the quadratic equation \( x^2 - 2kx - 6 = 0 \), then find the value of \( k \). 
Answer: Let \( \alpha \) be other root.
Product = \( \frac{c}{a} = \frac{-6}{1} = -6 \)

\( \implies \) \( 3 \times \alpha = -6 \)

\( \implies \) \( \alpha = -2 \).
Sum = \( \frac{-b}{a} = \frac{-(-2k)}{1} = 2k \).

\( \implies \) \( 3 + (-2) = 2k \)

\( \implies \) \( 1 = 2k, k = \frac{1}{2} \)
Value of \( k \) is \( \frac{1}{2} \).

Question. Write the discriminant of the quadratic equation: \( (x + 5)^2 = 2(5x - 3) \). 
Answer: \( (x + 5)^2 = 2(5x - 3) \)

\( \implies \) \( x^2 + 25 + 10x = 10x - 6 \)

\( \implies \) \( x^2 + 31 = 0 \).
\( a = 1, b = 0, c = 31 \)
Discriminant = \( b^2 - 4ac \)
\( = 0^2 - 4 \times 1 \times 31 \)
\( = 0 - 124 \)
\( = -124 \)
 

Question. If \( a \) and \( b \) are the roots of the equation \( x^2 + ax - b = 0 \), then find \( a \) and \( b \).
Answer: Sol. Sum of the roots = \( a + b = -\frac{B}{A} = -a \)
Product of the roots = \( ab = \frac{C}{A} = -b \)
\( \implies \) \( a + b = -a \) and \( ab = -b \)
\( \implies \) \( 2a = -b \) and \( a = -1 \)
\( \implies \) \( b = 2 \) and \( a = -1 \)

Question. Find the value of \( k \) for which the roots of the equation \( 3x^2 - 10x + k = 0 \) are reciprocal of each other? 
Answer: Sol. Given quadratic equation be
\( 3x^2 - 10x + k = 0 \)
Since its roots are reciprocal to each other.
\(\therefore\) Let one root be \(\alpha\), therefore other will be \(\frac{1}{\alpha}\).
Now, product of roots = \(\frac{c}{a}\).
\( \implies \) \(\alpha \times \frac{1}{\alpha} = \frac{k}{3}\)
\( \implies \) \(1 = \frac{k}{3}\)
\( \implies \) \(k = 3\)

Question. For what values of \( k \) does the quadratic equation \( 4x^2 - 12x - k = 0 \) have no real roots? 
Answer: Sol. For non-real roots, \( D < 0 \)
\( \implies \) \(144 - 4 \times 4 \times (-k) < 0\)
\(16k < -144\)
\(k < -9\)
[CBSE Marking Scheme 2019 (30/4/1)]

Short Answer Questions-

Question. Solve for \( x: 6x^2 + 11x + 3 = 0 \) 
Answer: Sol. Given equation,
\( 6x^2 + 11x + 3 = 0 \)
\( \implies \) \( 6x^2 + 9x + 2x + 3 = 0 \)
\( \implies \) \( 3x(2x + 3) + 1(2x + 3) = 0 \)
\( \implies \) \( (2x + 3)(3x + 1) = 0 \)
\( \implies \) \( 2x + 3 = 0 \) or \( 3x + 1 = 0 \)
\( \implies \) \( x = \frac{-3}{2}, \frac{-1}{3} \)

Question. Solve for \( x: \sqrt{2x + 9} + x = 13 \) 
Answer: Sol. \(\sqrt{2x+9} + x = 13\)
\(\sqrt{2x+9} = 13 - x\)
\(2x+9 = (13-x)^2\)
\( \implies \) \(2x+9 = 169 + x^2 - 26x\)
\( \implies \) \(x^2 + 169 - 26x - 9 - 2x = 0\)
\(x^2 - 28x + 160 = 0\)
\(x^2 - 20x - 8x + 160 = 0\)
\(x(x - 20) - 8(x - 20) = 0\)
\((x - 8)(x - 20) = 0\)
either \(x = 8\) or \(x = 20\)
[Topper's Answer 2016]

Question. If \( x = \frac{2}{3} \) and \( x = -3 \) are roots of the quadratic equation \( ax^2 + 7x + b = 0 \), find the values of \( a \) and \( b \).
Answer: Sol. Let us assume the quadratic equation be \( Ax^2 + Bx + C = 0 \).
Sum of the roots \( = -\frac{B}{A} \)
\( \implies \) \( \frac{-7}{a} = \frac{2}{3} - 3 \implies a = 3 \)
Product of the roots \( = \frac{C}{A} \)
\( \implies \) \( \frac{b}{a} = \frac{2}{3} \times (-3) \)
\( \implies \) \( \frac{b}{a} = -2 \)
\( \implies \) \( b = 3 \times (-2) \)
\( \implies \) \( b = -6 \)

Question. Find the roots of the quadratic equation \( -x^2 + 7x - 10 = 0 \) by using quadratic formula. 
Answer: Sol. Given quadratic equation
\( -x^2 + 7x - 10 = 0 \)
Its Discriminant \( (D) = b^2 - 4ac \)
\( \implies \) \( D = (7)^2 - 4 \times (-1) \times (-10) = 9 \)
\(\therefore\) \( x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-7 \pm \sqrt{9}}{2 \times (-1)} = \frac{-7 \pm 3}{-2} \)
\( \implies \) \( x = \frac{-7+3}{-2}, \frac{-7-3}{-2} = \frac{-4}{-2}, \frac{-10}{-2} \)
\( \implies \) \( x = 2, 5 \)
\(\therefore\) Roots are 2 and 5.

Question. Solve the quadratic equation \( 2x^2 + ax - a^2 = 0 \) for \( x \). 
Answer: Sol. \( 2x^2 + ax - a^2 = 0 \)
Here, \( a = 2, b = a \) and \( c = -a^2 \)
Using the formula,
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we get
\( x = \frac{-a \pm \sqrt{a^2 - 4 \times 2 \times (-a^2)}}{2 \times 2} = \frac{-a \pm \sqrt{9a^2}}{4} = \frac{-a \pm 3a}{4} \)
\( x = \frac{-a + 3a}{4} = \frac{a}{2}, x = \frac{-a - 3a}{4} = -a \implies x = \frac{a}{2}, -a \)

Question. Does there exist a quadratic equation whose co-efficients are rational but both of its roots are irrational? Justify your answer.
Answer: Sol. Yes, \( x^2 - 4x + 1 = 0 \) is a quadratic equation with rational co-efficients.
Its roots are \( \frac{4 \pm \sqrt{(-4)^2 - 4 \times 1 \times 1}}{2} = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3} \), which are irrational.

Question. Find the value of \( k \) for which the equation \( x^2 + k(2x + k - 1) + 2 = 0 \) has real and equal roots. 
Answer: Sol. Given quadratic equation:
\( x^2 + k(2x + k - 1) + 2 = 0 \)
\( \implies \) \( x^2 + 2kx + (k^2 - k + 2) = 0 \)
For equal roots, \( b^2 - 4ac = 0 \)
\( \implies \) \( 4k^2 - 4k^2 + 4k - 8 = 0 \)
\( \implies \) \( 4k = 8 \implies k = 2 \)