CBSE Class 10 Maths HOTs Quadratic Equations Set K

Very Short Answer Type Questions

Question. In a cricket match, Harbhajan took three wickets less than twice the number of wickets taken by Zahir. The Product of the number of wickets taken by these two is 20. Represent the above situation in the form of quadratic equation.
Answer: Let the number of wickets is taken by Zahir be \(x\), then number of wickets taken by Harbhajan will be \(2x - 3\).
According to question,
\(x(2x - 3) = 20\)
\(2x^2 - 3x = 20\)
Thus required quadratic equation,
\(2x^2 - 3x - 20 = 0\)

Question. Find the roots of \( x^2 - 4x - 8 = 0 \) by the method of completing square.
Answer: We have \( x^2 - 4x - 8 = 0 \)
Squaring both side we have
\( (x - 2)^2 - 8 - 4 = 0 \)
\( (x - 2)^2 - 12 = 0 \)
\( (x - 2)^2 = 12 \)
\( (x - 2)^2 = (2\sqrt{3})^2 \)
\( x - 2 = \pm 2\sqrt{3} \)
\( x = 2 \pm 2\sqrt{3} \)
Thus \( x = 2 + 2\sqrt{3}, 2 - 2\sqrt{3} \)

Question. Solve for \(x : \frac{1}{x} + \frac{2}{2x-3} = \frac{1}{x-2}, x \neq 0, \frac{2}{3}, 2.\)
Answer: We have \(\frac{1}{x} + \frac{2}{2x-3} = \frac{1}{x-2}\)
\(\frac{2x - 3 + 2x}{x(2x-3)} = \frac{1}{x-2}\)
\(\frac{4x - 3}{x(2x-3)} = \frac{1}{x-2}\)
\((x - 2)(4x - 3) = 2x^2 - 3x\)
\(4x^2 - 11x + 6 = 2x^2 - 3x\)
\(2x^2 - 8x + 6 = 0\)
\(x^2 - 4x + 3 = 0\)
\((x - 1)(x - 3) = 0\)
Thus \(x = 1, 3\)

Question. Solve the following quadratic equation for \(x : x^2 + \left( \frac{a}{a+b} + \frac{a+b}{a} \right)x + 1 = 0\)
Answer: We have \(x^2 + \left( \frac{a}{a+b} + \frac{a+b}{a} \right)x + 1 = 0\)
\(x^2 + \frac{a}{a+b}x + \frac{a+b}{a}x + 1 = 0\)
\(x \left( x + \frac{a}{a+b} \right) + \frac{a+b}{a} \left( x + \frac{a}{a+b} \right) = 0\)
\(\left( x + \frac{a}{a+b} \right) \left( x + \frac{a+b}{a} \right) = 0\)
Thus \(x = \frac{-a}{a+b}, \frac{-(a+b)}{a}\)

Question. Solve for \(x : \frac{1}{(x-1)(x-2)} + \frac{1}{(x-2)(x-3)} = \frac{2}{3}; x \neq 1, 2, 3\)
Answer: We have \(\frac{1}{(x-1)(x-2)} + \frac{1}{(x-2)(x-3)} = \frac{2}{3}\)
\(\frac{x-3+x-1}{(x-1)(x-2)(x-3)} = \frac{2}{3}\)
\(\frac{2x-4}{(x-1)(x-2)(x-3)} = \frac{2}{3}\)
\(\frac{2(x-2)}{(x-1)(x-2)(x-3)} = \frac{2}{3}\)
\(\frac{2}{(x-1)(x-3)} = \frac{2}{3}\)
\(3 = (x-1)(x-3)\)
\(x^2 - 4x + 3 = 3\)
\(x^2 - 4x = 0\)
\(x(x - 4) = 0\)
Thus \(x = 0\) or \(x = 4\)

Question. Solve for \(x : \sqrt{3}x^2 - 2\sqrt{2}x - 2\sqrt{3} = 0\)
Answer: We have \(\sqrt{3}x^2 - 2\sqrt{2}x - 2\sqrt{3} = 0\)
\(\sqrt{3}x^2 - [3\sqrt{2} - \sqrt{2}]x - 2\sqrt{3} = 0\)
\(\sqrt{3}x^2 - 3\sqrt{2}x + \sqrt{2}x - 2\sqrt{3} = 0\)
\(\sqrt{3}x^2 - \sqrt{3} \cdot \sqrt{3} \cdot \sqrt{2}x + \sqrt{2}x - \sqrt{2} \cdot \sqrt{2} \cdot \sqrt{3} = 0\)
\(\sqrt{3}x(x - \sqrt{6}) + \sqrt{2}(x - \sqrt{6}) = 0\)
\((x - \sqrt{6})(\sqrt{3}x + \sqrt{2}) = 0\)
Thus \(x = \sqrt{6}, -\sqrt{\frac{2}{3}}\)

Question. Solve for \(x : 2x^2 + 6\sqrt{3}x - 60 = 0\)
Answer: We have \(2x^2 + 6\sqrt{3}x - 60 = 0\)
\(x^2 + 3\sqrt{3}x - 30 = 0\)
\(x^2 + 5\sqrt{3}x - 2\sqrt{3}x - 30 = 0\)
\(x(x + 5\sqrt{3}) - 2\sqrt{3}(x + 5\sqrt{3}) = 0\)
\((x + 5\sqrt{3})(x - 2\sqrt{3}) = 0\)
Thus \(x = -5\sqrt{3}, 2\sqrt{3}\)

Question. Solve for \(x : \frac{16}{x} - 1 = \frac{15}{x+1}; x \neq 0, -1\)
Answer: We have \(\frac{16}{x} - 1 = \frac{15}{x+1}\)
\(\frac{16}{x} - \frac{15}{x+1} = 1\)
\(16(x + 1) - 15x = x^2 + x\)
\(16x + 16 - 15x = x^2 + x\)
\(x + 16 = x^2 + x\)
\(x^2 - 16 = 0\)
\(x^2 = 16\)
\(x = \pm 4\)
Thus \(x = -4\) and \(x = +4\)

Question. Solve the quadratic equation \((x - 1)^2 - 5(x - 1) - 6 = 0\)
Answer: We have \( (x - 1)^2 - 5(x - 1) - 6 = 0 \)
\( x^2 - 2x + 1 - 5x + 5 - 6 = 0 \)
\( x^2 - 7x + 6 - 6 = 0 \)
\( x^2 - 7x = 0 \)
\( x(x - 7) = 0 \)
Thus \( x = 0, 7 \)

Question. Solve for \( x \): \( \sqrt{3}x^2 - 2\sqrt{2}x - 2\sqrt{3} = 0 \)
Answer: We have \( \sqrt{3}x^2 - 3\sqrt{2}x + \sqrt{2}x - 2\sqrt{3} = 0 \)
\( \sqrt{3}x[x - \sqrt{6}] + \sqrt{2}[x - \sqrt{6}] = 0 \)
\( (x - \sqrt{6})(\sqrt{3}x + \sqrt{2}) = 0 \)
Thus \( x = \sqrt{6}, -\sqrt{\frac{2}{3}} \)

Short Answer Type Questions

Question. If the roots of the quadratic equation \((x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0\) are equal. Then show that \(a = b = c\).
Answer: We have \((x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0\)
\(x^2 - ax - bx + ab + x^2 - bx - cx + bc + x^2 - cx - ax + ac = 0\)
\(3x^2 - 2(a + b + c)x + ab + bc + ca = 0\)
For equal roots \(B^2 - 4AC = 0\)
\(\{-2(a + b + c)\}^2 - 4 \times 3(ab + bc + ca) = 0\)
\(4(a + b + c)^2 - 12(ab + bc + ca) = 0\)
\((a + b + c)^2 - 3(ab + bc + ca) = 0\)
\(a^2 + b^2 + c^2 + 2ab + 2bc + 2ac - 3ab - 3bc - 3ac = 0\)
\(a^2 + b^2 + c^2 - ab - bc - ac = 0\)
\(\frac{1}{2}[2a^2 + 2b^2 + 2c^2 - 2ab - 2ac - 2bc] = 0\)
\(\frac{1}{2}[(a^2 + b^2 - 2ab) + (b^2 + c^2 - 2bc) + (c^2 + a^2 - 2ac)] = 0\)
or, \((a - b)^2 + (b - c)^2 + (c - a)^2 = 0\)
If \(a \neq b \neq c\)
\((a - b)^2 > 0, (b - c)^2 > 0, (c - a)^2 > 0\)
If \((a - b)^2 = 0 \Rightarrow a = b\)
\((b - c)^2 = 0 \Rightarrow b = c\)
\((c - a)^2 = 0 \Rightarrow c = a\)
Thus \(a = b = c\)
Hence Proved

Question. If the roots of the quadratic equation \((c^2 - ab)x^2 - 2(a^2 - bc)x + b^2 - ac = 0\) in \(x\) are equal then show that either \(a = 0\) or \(a^3 + b^3 + c^3 = 3abc\)
Answer: We have \((c^2 - ab)x^2 - 2(a^2 - bc)x + b^2 - ac = 0\)
Compare with \(Ax^2 + Bx + C = 0\) we get
\(A = (c^2 - ab), B = -2(a^2 - bc), C = (b^2 - ac)\)
If roots are equal, \(B^2 - 4AC = 0\)
\([2(a^2 - bc)]^2 - 4(c^2 - ab)(b^2 - ac) = 0\)
\(4[a^4 + b^2c^2 - 2a^2bc] - 4(b^2c^2 - ac^3 - ab^3 + a^2bc) = 0\)
\(4[a^4 + b^2c^2 - 2a^2bc - b^2c^2 + ac^3 + ab^3 - a^2bc] = 0\)
\(4[a^4 + ac^3 + ab^3 - 3a^2bc] = 0\)
\(a(a^3 + c^3 + b^3 - 3abc) = 0\)
\(a = 0\) or \(a^3 + b^3 + c^3 = 3abc\)

Question. Solve for \(x : \frac{1}{a + b + x} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x}\) where \(a + b + x \neq 0\) and \(a, b, x \neq 0\)
Answer: We have \(\frac{1}{a + b + x} - \frac{1}{x} = \frac{1}{a} + \frac{1}{b}\)
\(\frac{-(a + b)}{x^2 + (a + b)x} = \frac{b + a}{ab}\)
\(x^2 + (a + b)x + ab = 0\)
\((x + a)(x + b) = 0\)
\(x = -a, x = -b\)
Hence \(x = -a, -b\)

Question. The difference between the radii of the smaller circle and the larger circle is 7 cm and the difference between their areas of the two circles in 1078 sq. cm. Find the radius of the smaller circle.
Answer: We have \( r_2 - r_1 = 7 \) cm, \( r_2 > r_1 \) ...(1)
and \( \pi(r_2^2 - r_1^2) = 1078 \) cm²
\( \pi(r_2 - r_1)(r_2 + r_1) = 1078 \)
\( \frac{22}{7} \times 7(r_2 + r_1) = 1078 \)
\( r_2 + r_1 = \frac{1078}{22} = 49 \) ...(2)
Adding (1) and (2) we get
\( 2r_2 = 56 \)
\( r_2 = 28 \) cm
and \( r_1 = 49 - 28 = 21 \)
Hence radii of two circles are 28 cm and 21 cm.

Question. A train travelling at a uniform speed for 360 km have taken 48 minutes less to travel the same distance if its speed were 5 km/hour more. Find the original speed of the train.
Answer: Let the original speed of the train be \( x \) km/hr.
Time taken = \( \frac{Distance}{Speed} = \frac{360}{x} \) hours
Time taken at increase speed = \( \frac{360}{x+5} \) hours.
According to the question
\( \frac{360}{x} - \frac{360}{x+5} = \frac{48}{60} \)
\( 360 [\frac{1}{x} - \frac{1}{x+5}] = \frac{4}{5} \)
\( \frac{1800}{x^2+5x} = \frac{4}{5} \)
\( x^2 + 5x - 2250 = 0 \)
\( x^2 + (50 - 45)x - 2250 = 0 \)
\( x^2 + 50x - 45x - 2250 = 0 \)
\( (x + 50)(x - 45) = 0 \)
\( x = -50 \) or \( x = 45 \)
As speed can not be negative, original speed of train is 45 km/hr.

Question. Check whether the equation \( 5x^2 - 6x - 2 = 0 \) has real roots if it has, find them by the method of completing the square. Also verify that roots obtained satisfy the given equation.
Answer: We have \( 5x^2 - 6x - 2 = 0 \)
Compare with \( ax^2 + bx + c = 0 \) we get
\( a = 5, b = (-6) \) and \( c = (-2) \)
\( b^2 - 4ac = (-6)^2 - 4 \times 5 \times (-2) \)
\( = 36 + 40 = 76 > 0 \)
So the equation has real and two distinct roots.
\( 5x^2 - 6x = 2 \)
Dividing both the sides by 5 we get
\( x^2 - \frac{6}{5}x = \frac{2}{5} \)
\( x^2 - 2x(\frac{3}{5}) = \frac{2}{5} \)
Adding square of the half of coefficient of \( x \)
\( x^2 - 2x(\frac{3}{5}) + \frac{9}{25} = \frac{2}{5} + \frac{9}{25} \)
\( (x - \frac{3}{5})^2 = \frac{19}{25} \)
\( x - \frac{3}{5} = \pm \frac{\sqrt{19}}{5} \)
\( x = \frac{3 + \sqrt{19}}{5} \) or \( \frac{3 - \sqrt{19}}{5} \)
Verification :
\( 5[\frac{3+\sqrt{19}}{5}]^2 - 6[\frac{3+\sqrt{19}}{5}] - 2 \)
\( = 9 + \frac{6\sqrt{19}+19}{5} - (\frac{18 + 6\sqrt{19}}{5}) - 2 \)
\( = \frac{28+6\sqrt{19}}{5} - \frac{18+6\sqrt{19}}{5} - 2 \)
\( = \frac{28+6\sqrt{19}-18-6\sqrt{19}-10}{5} \)
\( = 0 \)
Similarly \( 5[\frac{3-\sqrt{19}}{5}]^2 - 6[\frac{3-\sqrt{19}}{5}] - 2 = 0 \)
Hence verified.

HOTS QUESTIONS

Question. Solve \( \frac{1}{(a+b+x)} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x}, a, b \neq 0 \).
Answer: We have \( \frac{1}{a+b+x} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x} \)
\( \frac{1}{a+b+x} - \frac{1}{x} = \frac{1}{a} + \frac{1}{b} \)
\( \frac{x-(a+b+x)}{x(a+b+x)} = \frac{a+b}{ab} \)
\( \frac{x-a-b-x}{x(a+b+x)} = \frac{a+b}{ab} \)
\( \frac{-(a+b)}{x(a+b+x)} = \frac{a+b}{ab} \)
\( x(a+b+x) = -ab \)
\( x^2 + (a+b)x + ab = 0 \)
\( (x+a)(x+b) = 0 \)
\( x = -a \) or \( x = -b \)

Question. Find the value of k for which the distance between (9,2) and (3,k) is 10 units.
Answer: \( d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} \)
Thus \( \sqrt{(3-9)^2 + (k-2)^2} = 10 \)
\( (-6)^2 + k^2 - 4k + 4 = 100 \)
\( k^2 - 4k + 40 = 100 \)
\( k^2 - 4k - 60 = 0 \)
\( k^2 - 10k + 6k - 60 = 0 \)
\( k(k-10) + 6(k-10) = 0 \)
\( (k-10)(k+6) = 0 \)
\( k = 10, -6 \)

Question. A shopkeeper buys a number of books for Rs. 1200. If he had bought 10 more books for the same amount, each book would have cost him Rs. 20 less. How many books did he buy?
Answer: Let the number of books bought be \( x \).
As per question we have
\( \frac{1200}{x} - \frac{1200}{x+10} = 20 \)
\( x^2 + 10x - 600 = 0 \)
\( (x+30)(x-20) = 0 \)
\( x = -30 \) or \( x = 20 \)
Since number of books cannot be negative, \( x = 20 \)
Thus number of books bought is 20.

Question. If the price of a book is reduced by Rs. 5, a person can buy 5 more book for Rs. 300. Find the original list price of the book.
Answer: Let the original list price be Rs. \( x \)
No. of books bought for Rs. 300 = \( \frac{300}{x} \)
Reduced list price of the book = \( (x - 5) \)
No. of books brought for Rs. 300 = \( \frac{300}{x-5} \)
According to questions, we have
\( \frac{300}{x-5} - \frac{300}{x} = 5 \)
\( x^2 - 5x - 300 = 0 \)
\( (x - 20)(x + 15) = 0 \)
Since price cannot be negative, \( x = 20 \)
Thus original list price is 20 rs.

Question. A car covers a distance of 2592 km with a uniform speed. The number of hours taken for journey is one half the number representing the speed in km/hour. Find the time taken to cover the distances.
Answer: Let the speed of the car be \( x \) km/hr.
Therefore time taken \( = \frac{x}{2} \) hour
Now Speed \( = \frac{\text{Distance}}{\text{Time}} \)
\( x = \frac{2592}{\frac{x}{2}} \)
\( x^2 = 2592 \times 2 = 5184 \)
\( x = \sqrt{5184} = 72 \)
Hence the time taken \( \frac{72}{2} = 36 \) hours.

Question. Speed of a boat in still water is 15 km/hour. It goes 30 km up stream and returns back at the same point in 4 hours 30 minutes. Find the speed of the stream.
Answer: Let the speed of the Stream be \( x \) km/hr.
Speed of boat up stream \( = 15 - x \)
and speed of boat down stream \( = 15 + x \)
According to the question
\( \frac{30}{15 - x} + \frac{30}{15 + x} = 4 \frac{1}{2} \)
\( \frac{30(15 + x) + 30(15 - x)}{15^2 - x^2} = \frac{9}{2} \)
\( 900 \times 2 = 9(15^2 - x^2) \)
\( 9x^2 = 2025 - 1800 = 225 \)
\( x^2 = 25 \)
\( x = \pm 5 \)
Hence, the speed of the stream \( = 5 \) km/hr

Question. A takes 6 days less than B to do a work. If both A and B working together can do it in 4 days, how many days will B take to finish it ?
Answer: Let B complete a work in \( x \) days.
Then A takes \( x - 6 \) days to complete it.
Together they complete it in 4 days.
According to work done per day,
\( \frac{1}{x - 6} + \frac{1}{x} = \frac{1}{4} \)
\( \frac{x + x - 6}{x(x - 6)} = \frac{1}{4} \)
\( 4(2x - 6) = x(x - 6) \)
\( 8x - 24 = x^2 - 6x \)
\( x^2 - 14x + 24 = 0 \)
\( x^2 - 12x - 2x + 24 = 0 \)
\( x(x - 12) - 2(x - 12) = 0 \)
\( (x - 2)(x - 12) = 0 \)
\( x = 2 \) or \( 12 \)
Here \( x = 2 \) is not possible because \( x - 6 \) is \( (-4) \) which is not possible.
Thus \( x = 12 \) and B takes 12 days to finish the work.

Question. In a class test Raveena got a total of 30 mark in English and Mathematics. Had she got 2 more marks in Mathematics and 3 marks less in English then the product of her marks obtained would have be 210. Find the individual marks obtained in two subjects.
Answer: Let Raveena got marks in English be \( x \).
Marks in Mathematics \( = (30 - x) \)
According to problem
\( (x - 3)((30 - x) + 2) = 210 \)
\( (x - 3)(32 - x) = 210 \)
\( 35x - 96 - x^2 = 210 \)
\( x^2 - 35x + 306 = 0 \)
\( x^2 - 18x - 17x + 306 = 0 \)
\( x(x - 18) - 17(x - 18) = 0 \)
\( (x - 18)(x - 17) = 0 \)
\( x = 18, 17 \)
Hence, if she got 18 marks in English, then she got 12 in mathematics.
If she got 17 marks in English, then she got 13 marks in mathematics.

Question. Two taps running together can fill a tank in \( 3 \frac{1}{13} \) hours. If one tap takes 3 hours more than the other to fill the tank, then how much time will each tap take to fill the tank?
Answer: Two tap running together fill the tank in \( 3 \frac{1}{13} \) hr.
\( = \frac{40}{13} \) hours
Thus It will fill in 1 hour \( \frac{13}{40} \) tank
If first tap alone fills the tank in \( x \) hrs, then second tap alone fills it in \( (x + 3) \) hr.
Now, \( \frac{1}{x} + \frac{1}{x + 3} = \frac{13}{40} \)
\( \frac{x + 3 + x}{x(x + 3)} = \frac{13}{40} \)
\( 80x + 120 = 13x^2 + 39x \)
or, \( 13x^2 - 41x - 120 = 0 \)
\( 13x^2 - (65 - 24)x - 120 = 0 \)
\( (x - 5)(13x + 24) = 0 \)
\( x = 5, x = -\frac{24}{13} \)
Here \( x = -\frac{24}{13} \) is not possible. Hence, 1st tap takes 5 hours and 2nd tap takes \( 5 + 3 = 8 \) hours.

Question. Two taps running together can fill a cistern in \( 2 \frac{8}{11} \) minutes. If one tap takes 1 minute more than the other to fill the cistern, find the time in which each tap separately can fill the cistern.
Answer: Two taps together fill the cistern in \( 2 \frac{8}{11} \) minutes
\( = \frac{30}{11} \) minutes
Thus It will fill in 1 minute \( \frac{11}{30} \) cistern
Let first tap fills the same cistern in \( x \) minutes and 2nd tap will take \( = (x + 1) \) minutes
Thus \( \frac{1}{x} + \frac{1}{x + 1} = \frac{11}{30} \)
\( \frac{x + 1 + x}{x(x + 1)} = \frac{11}{30} \)
\( 60x + 30 = 11x^2 + 11x \)
\( 11x^2 - 49x - 30 = 0 \)
\( 11x^2 - (55 - 6)x - 30 = 0 \)
\( 11x^2 - 55x + 6x - 30 = 0 \)
\( 11x(x - 5) + 6(x - 5) = 0 \)
\( (x - 5)(11x + 6) = 0 \)
\( x = 5, x = -\frac{6}{11} \)
Here \( x = -\frac{6}{11} \) is not possible. Hence, 1st tap takes 5 minute and 2nd tap takes 6 minute.

Question. A and B working together can do a work in 6 days. If A takes 5 days less than B to finish the work, in how many days B alone can do it alone?
Answer: Since \( A + B \) finish the work in 6 days.
They will finish in one day \( \frac{1}{6} \) work
Let B alone does the same work in \( x \) days, then A alone will finish in \( (x - 5) \) days.
Now, \( \frac{1}{x - 5} + \frac{1}{x} = \frac{1}{6} \)
\( \frac{x + x - 5}{x(x - 5)} = \frac{1}{6} \)
\( 12x - 30 = x^2 - 5x \)
\( x^2 - 17x + 30 = 0 \)
\( x^2 - 15x - 2x + 30 = 0 \)
\( x(x - 15) - 2(x - 15) = 0 \)
\( (x - 15)(x - 2) = 0 \)
\( x = 15, x = 2 \)
Here \( x = 2 \) is not possible. Hence, B finishes the work in 15 days.

Question. Ram takes 6 days less than Bhagat to finish a place of work. If both of them together can finish the work in 4 days, in how many days Bhagat alone can finish the work ?
Answer: Ram and Bhagat together do the work in 4 days
Ram and Bhagat will do in one days \( \frac{1}{4} \) work
Let Bhagat alone does the same work in \( x \) days.
Ram will take \( = (x - 6) \) days
Now \( \frac{1}{x} + \frac{1}{x - 6} = \frac{1}{4} \)
\( \frac{x - 6 + x}{x(x - 6)} = \frac{1}{4} \)
\( 8x - 24 = x^2 - 6x \)
\( x^2 - 14x + 24 = 0 \)
\( x^2 - 12x - 2x + 24 = 0 \)
\( x(x - 12) - 2(x - 12) = 0 \)
\( (x - 12)(x - 2) = 0 \)
\( x = 12, x = 2 \)
If Bhagat complete the work in 2 days, Ram will take \( = 2 - 6 = -4 \) days that is impossible. Hence, Bhagat can finish in 12 days.