CBSE Class 10 Mathematics Quadratic Equations Worksheet Set 06

Question. \( 2x^2 - 3x + 2 = 0 \) have 
(a) Real and Distinct roots
(b) Real and Equal roots
(c) Real roots
(d) No Real roots
Answer: (d) No Real roots

Question. \( 2x^2 + 5\sqrt{3}x + 6 = 0 \) have 
(a) Real and equal root
(b) Real roots
(c) No Real roots
(d) Real and Distinct roots
Answer: (d) Real and Distinct roots

Question. If one root of the equation \( 4x^2 - 2x + (\lambda - 4) = 0 \) be the reciprocal of the other, then the value of \( \lambda \) is 
(a) 8
(b) 4
(c) - 8
(d) - 4
Answer: (a) 8

Question. A quadratic equation \( ax^2 + bx + c = 0 \) has non – real roots, if 
(a) \( b^2 - 4ac > 0 \)
(b) \( b^2 - 4ac = 0 \)
(c) \( b^2 - 4ac < 0 \)
(d) \( b^2 - ac = 0 \)
Answer: (c) \( b^2 - 4ac < 0 \)

Question. The quadratic equation whose roots are \( \frac{-1}{3} \) and \( \frac{5}{2} \) is 
(a) \( 6x^2 + 13x - 5 = 0 \)
(b) \( 6x^2 + 13x + 5 = 0 \)
(c) \( 6x^2 - 13x + 5 = 0 \)
(d) \( 6x^2 - 13x - 5 = 0 \)
Answer: (d) \( 6x^2 - 13x - 5 = 0 \)

Question. What is the nature of roots of the quadratic equation \( 5x^2 - 2x - 3 = 0 \)? 
Answer: On comparing equation with standard form of equation i.e, \( ax^2 + bx + c = 0 \), we get
a = 5, b = -2, c = -3
Now , \( D = b^2 - 4ac = (-2)^2 - 4 \times 5 \times (-3) \)
= 4 + 60 = 64
Therefore, D = 64
We know , For D>0 , the roots of equation are real and distinct.
Therefore , \( 5x^2 - 2x - 3 = 0 \) has real and distinct roots.

Question. If \( x = \frac{-1}{2} \), is a solution of the quadratic equation \( 3x^2 + 2kx - 3 = 0 \), find the value of k. 
Answer: We have,
\( 3x^2 +2kx - 3 = 0 \)
since \( x = \frac{-1}{2} \)
\( \therefore 3(\frac{-1}{2})^2 + 2k(\frac{-1}{2}) - 3 = 0 \)
\( \Rightarrow 3(\frac{1}{4}) - k - 3 = 0 \)
\( \Rightarrow \frac{3}{4} - k - 3 = 0 \)
\( \Rightarrow k = \frac{3}{4} - 3 = \frac{3-12}{4} = - \frac{9}{4} \)

Question. Check whether \( (x - 7)x = 3x^2 - 5 \) is a quadratic equation: 
Answer: Given equation is \( (x - 7)x = 3x^2 - 5 \)
\( \Rightarrow x^2 - 7x = 3x^2 - 5 \)
\( \Rightarrow 2x^2 + 7x - 5 = 0 \)
which of the form \( ax^2 + bx + c = 0 \)
Hence, given equation is a quadratic equation.

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. Solve the following problem: \( x^2 - 55x + 750 = 0 \) 
Answer: \( x^2 - 55x + 750 = 0 \)
\( \Rightarrow x^2 - 25x - 30x + 750 = 0 \Rightarrow x(x - 25) - 30(x - 25) = 0 \)
\( \Rightarrow (x - 30)(x - 25) \Rightarrow x = 30, 25 \)

Question. Solve the quadratic equation \( x^2 + \left( \frac{a}{a+b} + \frac{a+b}{a} \right)x + 1 = 0 \) by factorization. 
Answer: We have, \( x^2 + \left( \frac{a}{a+b} + \frac{a+b}{a} \right)x + 1 = 0 \)
\( \Longrightarrow x^2 + \frac{a}{a+b}x + \frac{a+b}{a}x + 1 = 0 \)
\( \Longrightarrow x(x + \frac{a}{a+b}) + \frac{a+b}{a}(x + \frac{a}{a+b}) = 0 \)
\( \Longrightarrow (x + \frac{a}{a+b})(x + \frac{a+b}{a}) = 0 \)
Either \( x + \frac{a}{a+b} = 0 \) or \( x + \frac{a+b}{a} = 0 \)
\( \Longrightarrow x = - \frac{a}{a+b}, - \frac{a+b}{a} \)
\( \therefore x = - \frac{a}{a+b}, - \frac{a+b}{a} \) are the required roots.

Question. A water tank is connected with three pipes of uniform flow. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. If the fist two pipes operating simultaneously, fill the tank in the same time as the time taken by the third pipe alone to fill the tank. Find the time taken by each pipe individually to fill the tank. 
Answer: Let the time taken by the second pipe to fill the tank individually be x hrs.
Then, the time taken by the first and third pipes to fill the tank individually are (x + 5) hrs and (x – 4) hrs respectively.
We have, \( \frac{1}{x+5} + \frac{1}{x} = \frac{1}{x-4} \)
\( \Longrightarrow \frac{x+x+5}{x(x+5)} = \frac{1}{x-4} \)
\( \Longrightarrow (2x + 5)(x - 4) = x^2 + 5x \)
\( \Longrightarrow 2x^2 - 8x + 5x - 20 = x^2 + 5x \)
\( \Longrightarrow x^2 - 8x - 20 = 0 \)
\( \Longrightarrow x^2 - 10x + 2x - 20 = 0 \)
\( \Longrightarrow x(x - 10) + 2(x - 10) = 0 \)
\( \Longrightarrow (x - 10)(x + 2) = 0 \)
Either x = 10 or x = - 2
But x cannot be negative. So, x = 10
Hence, the time taken by first, second and third pipe to fill the tank individually are 15 hours, 10 hours and 6 hours respectively.

Question. Determine whether the given values are solution of the given equation or not: \( 6x^2 - x - 2 = 0, x = -1/2, x = 2/3 \) 
Answer: We have, \( 6x^2 - x - 2 = 0 \)
Substituting \( x = - \frac{1}{2} \), we get
LHS = \( 6 \times (- \frac{1}{2})^2 - (- \frac{1}{2}) - 2 = \frac{6}{4} + \frac{1}{2} - 2 = 0 = \) RHS
Therefore, \( x = - \frac{1}{2} \) is a solution of the given equation.
Now, \( x = \frac{2}{3} \),
LHS = \( 6 \times (\frac{2}{3})^2 - \frac{2}{3} - 2 = 0 = \) RHS
Therefore, \( x = \frac{2}{3} \) is also a solution of the given equation.

Question. A farmer wishes to grow a \( 100 \text{ m}^2 \) rectangular vegetable garden. Since he has with him only 30 m barbed wire, he fences three sides of the rectangular garden letting compound wall of his house act as the fourth side-fence. Find the dimensions of his garden. 
Answer: Let the length of one side be x metres and other side be y metres.
Then, x + y + x = 30 \( \Longrightarrow \) y = 30 - 2x
\( \therefore \) Area of the garden = 100 \( \text{m}^2 \)
\( \Longrightarrow xy = 100 \)
\( \Longrightarrow x(30 - 2x) = 100 \)
\( \Longrightarrow 2x^2 - 30x + 100 = 0 \)
\( \Longrightarrow 2(x^2 - 15x + 50) = 0 \) or \( x^2 - 15x + 50 = 0 \)
\( \Longrightarrow x^2 - 10x - 5x + 50 = 0 \)
\( \Longrightarrow x(x - 10) - 5(x - 10) = 0 \)
\( \Longrightarrow (x - 10)(x - 5) = 0 \)
Either x - 10 = 0 or x - 5 = 0
x = 10, 5
\( \therefore y = 30 - 20 = 10 \) or \( 30 - 10 = 20 \)
Hence, the dimensions of the vegetable garden are 5m \( \times \) 20m or 10m \( \times \) 10m.

Question. Solve for x: \( \frac{2x}{x-3} + \frac{1}{2x+3} + \frac{3x+9}{(x-3)(2x+3)} = 0, x \neq 3, -\frac{3}{2} \) 
Answer: Given, \( \frac{2x}{x-3} + \frac{1}{2x+3} + \frac{3x+9}{(x-3)(2x+3)} = 0 \)
\( \Longrightarrow \frac{2x(2x+3) + (x-3) + 3x + 9}{(x-3)(2x+3)} = 0 \)
\( \Longrightarrow \frac{4x^2 + 6x + x - 3 + 3x + 9}{2x^2 + 3x - 6x - 9} = 0 \)
\( \Longrightarrow \frac{4x^2 + 10x + 6}{2x^2 - 3x - 9} = 0 \)
Cross multiplying equation, we get,
\( \Longrightarrow 4x^2 + 10x + 6 = 0 \)
\( \Longrightarrow 4x^2 + 6x + 4x + 6 = 0 \)
\( \Longrightarrow x(4x + 6) + 1(4x + 6) = 0 \)
\( \Longrightarrow (x + 1) = 0 \) or \( (4x + 6) = 0 \)
\( \Longrightarrow x = - 1 \)
\( \Longrightarrow x = - \frac{3}{2} \)
Hence , the roots of the given quadratic equation are -1 and \( - \frac{3}{2} \)

Question. A motor boat, whose speed is 15 km/hr in still water, goes 30 km downstream and comes back in a total time of 4 hr 30 minutes, find the speed of the stream. 
Answer: Speed of motor boat in still water = 15 km/hr
Speed of stream = x km/hr
Speed in downward direction = 15 + x
Speed in upward direction = 15 - x
According to question,
\( \frac{30}{15+x} + \frac{30}{15-x} = 4 \frac{1}{2} \)
\( \Longrightarrow \frac{30(15-x) + 30(15+x)}{(15+x)(15-x)} = \frac{9}{2} \)
\( \Longrightarrow \frac{450-30x+450+30x}{225-x^2} = \frac{9}{2} \)
\( \Longrightarrow 9(225 - x^2) = 1800 \)
\( \Longrightarrow 225 - x^2 = 200 \)
\( \Longrightarrow x = 5 \)
Speed of stream = 5 km/hr

Question. Solve for x by quadratic formula \( p^2x^2 + (p^2 - q^2)x - q^2 = 0 \) 
Answer: \( p^2x^2 + (p^2 - q^2)x - q^2 = 0 \)
\( a = p^2, b = p^2 - q^2, c = -q^2 \)
\( D = b^2 - 4ac \)
= \( (p^2 - q^2)^2 - 4 \times p^2(-q^2) \)
= \( p^4 + q^4 - 2p^2q^2 + 4p^2q^2 \)
= \( (p^2 + q^2)^2 \)
\( x = \frac{-b \pm \sqrt{D}}{2a} \)
= \( \frac{-(p^2 - q^2) \pm \sqrt{(p^2 + q^2)^2}}{2 \times p^2} = \frac{-p^2 + q^2 \pm (p^2 + q^2)}{2p^2} \)
or \( x = \frac{-p^2 + q^2 + p^2 + q^2}{2p^2} \)
\( x = \frac{2q^2}{2p^2} \) or \( x = \frac{-2p^2}{2p^2} \)
\( x = \frac{q^2}{p^2} \) or \( x = - 1 \)

Question. A train travels a distance of 480 km's at a uniform speed. If the speed had been 8 km/hr less, then it would have taken 3 hours more to cover the same distance. Find the quadratic equation in terms of the speed of the train. (4)
Answer: Distance travelled by the train = 480 km
Let the speed of the train be x kmph
Time taken for the journey = \( \frac{480}{x} \)
Given speed is decreased by 8 kmph
Hence the new speed of train = (x – 8) kmph
Time taken for the journey = \( \frac{480}{x-8} \)
\( \frac{480}{x-8} = \frac{480}{x} + 3 \)
\( \Longrightarrow \frac{480}{x-8} - \frac{480}{x} = 3 \)
\( \Longrightarrow \frac{480(x - x + 8)}{x(x-8)} = 3 \)
\( \Longrightarrow \frac{480 \times 8}{x(x-8)} = 3 \)
\( \Longrightarrow 3x(x - 8) = 480 \times 8 \)
\( \Longrightarrow x(x - 8) = 160 \times 8 \)
\( \Longrightarrow x^2 - 8x - 1280 = 0 \)
On solving we get x = 40
Thus the speed of train is 40 kmph.

Question. A journey of 192 km from a town A to town B takes 2 hours more by an ordinary passenger train than a super fast train. If the speed of the faster train is 16 km/h more, find the speed of the faster and the passenger train. 
Answer: Let speed of passenger train be x km/h
\( \therefore \) speed of superfast train = (x + 16) km/h
By question, \( T_{passenger} = \frac{192}{x} \) and \( T_{superfast} = \frac{192}{(x+16)} \)
or, \( \frac{192}{x} - \frac{192}{x+16} = 2 \)
or, \( 192(x + 16) - 192x = 2(x^2 + 16x) \)
or, \( 192x + 192 \times 16 - 192x = 2(x^2 + 16x) \)
\( 192x + 3072 - 192x = 2(x^2 + 16x) \) ( divide throughout by 2, we get,
\( 96x + 1536 - 96x = (x^2 + 16x) \)
or \( x(x + 48) - 32(x + 48) = 0 \)
or, \( (x - 32) (x + 48) = 0 \)
or, x = 32 or - 48
Since speed can't be negative, therefore - 48 is not possible.
\( \therefore \) Speed of passenger train = 32 km/h and Speed of fast train = 48 km/h

Question. Solve for x: \( \left( \frac{2x}{x-5} \right)^2 + 5 \left( \frac{2x}{x-5} \right) - 24 = 0, x \neq 5 \) 
Answer: We have given,
\( \left( \frac{2x}{x-5} \right)^2 + 5 \left( \frac{2x}{x-5} \right) - 24 = 0 \)
Let \( \frac{2x}{(x-5)} \) be y
\( \therefore y^2 + 5y - 24 = 0 \)
Now factorise,
\( y^2 + 8y - 3y - 24 = 0 \)
\( y(y + 8) - 3(y + 8) = 0 \)
\( (y + 8)(y - 3) = 0 \)
y = 3, -8
Putting y=3
\( \frac{2x}{x-5} = 3 \)
2x = 3x - 15
x = 15
Putting y = -8
\( \frac{2x}{x-5} = - 8 \)
2x = -8x + 40
10x = 40
x = 4
Hence, x is 15 , 4

Assertion-Reason Questions

The following questions consist of two statements—Assertion(A) and Reason(R). Answer these questions selecting the appropriate option given below:
(a) Both A and R are true and R is the correct explanation for A.
(b) Both A and R are true but R is not the correct explanation for A.
(c) A is true but R is false.
(d) A is false but R is true.

Question. Assertion (A) : The equation \( x^2 + 3x + 1 = (x - 2)^2 \) is a quadratic equation.
Reason (R) : Any equation of the form \( ax^2 + bx + c = 0 \) where \( a \neq 0 \), is called a quadratic equation.
Answer: Solution : We have, \( x^2 + 3x + 1 = (x - 2)^2 = x^2 - 4x + 4 \)

\( \implies x^2 + 3x + 1 = x^2 - 4x + 4 \)

\( \implies 7x - 3 = 0 \), it is not of the form \( ax^2 + bx + c = 0 \).
So, A is false but R is true.
Hence, option (d) is correct.

Question. Assertion (A) : The roots of the quadratic equation \( x^2 + 2x + 2 = 0 \) are not real.
Reason (R) : If discriminant \( D = b^2 - 4ac < 0 \) then the roots of quadratic equation \( ax^2 + bx + c = 0 \) are not real.
Answer: Solution : \( x^2 + 2x + 2 = 0 \)
\( \therefore \text{Discriminant, } D = b^2 - 4ac = (2)^2 - 4 \times 1 \times 2 = 4 - 8 = -4 < 0 \)
\( \therefore \text{Roots are not real.} \)
So, both A and R are true and R is the correct explanation for A.
Hence, option (a) is correct.

Question. Assertion (A) : The value of \( k = 2 \), if one root of the quadratic equation \( 6x^2 - x - k = 0 \) is \( \frac{2}{3} \).
Reason (R) : The quadratic equation \( ax^2 + bx + c = 0, a \neq 0 \) has two roots.
Answer: Solution : As one root is \( \frac{2}{3} \)

\( \implies x = \frac{2}{3} \)
\( \therefore 6 \times \left( \frac{2}{3} \right)^2 - \frac{2}{3} - k = 0 \)

\( \implies 6 \times \frac{4}{9} - \frac{2}{3} = k \)

\( \implies k = \frac{8}{3} - \frac{2}{3} = \frac{6}{3} = 2 \)

\( \implies k = 2 \)
So, both A and R are true but R is not the correct explanation for A.
Hence, option (b) is correct.

Question. Assertion (A) : If the equation \( 8x^2 + 3kx + 2 = 0 \) has equal roots then the value of \( k \) is \( \pm \frac{8}{3} \).
Reason (R) : The equation \( ax^2 + bx + c = 0 \) has equal roots if \( D = b^2 - 4ac = 0 \).
Answer: Solution : \( 8x^2 + 3kx + 2 = 0 \)
\( \therefore \text{Discriminant, } D = b^2 - 4ac = (3k)^2 - 4 \times 8 \times 2 = 9k^2 - 64 \)
For equal roots, \( D = 0 \)

\( \implies 9k^2 - 64 = 0 \)

\( \implies 9k^2 = 64 \)

\( \implies k^2 = \frac{64}{9} \)

\( \implies k = \pm \frac{8}{3} \)
So, A and R both are true and R is the correct explanation for A.
Hence, option (a) is correct.

Question. Assertion (A) : The values of \( x \) are \( -\frac{a}{2}, a \) for a quadratic equation \( 2x^2 + ax - a^2 = 0 \).
Reason (R) : For quadratic equation \( ax^2 + bx + c = 0 \), \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
Answer: Solution : \( 2x^2 + ax - a^2 = 0 \)
\( x = \frac{-a \pm \sqrt{a^2 + 8a^2}}{4} = \frac{-a \pm 3a}{4} = \frac{2a}{4}, \frac{-4a}{4} \)

\( \implies x = \frac{a}{2}, -a \)
So, A is false but R is true.
Hence, option (d) is correct.