CBSE Class 10 Quadratic Equations Sure Shot Questions Set 03

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Show that each of the following equations has real roots, and solve each by using the quadratic formula:

1. 9x²+7x-2=0

2. x²+6x+6=0

3. 2x²+5√ 3x+6=0

4. 36x²-12ax+(a²-b²)=0

5.(a+b)2x²-(4b⁴-3a⁴)x-12a²b²=0

6. (a+b)²x²-4abx-(a-b)²=0

7. 4x²-2(a²+b²)x+a²b²=0

8. 9x²-9(a+b)x+(2x²+5ab-2b²)=0

9. 4x²-4a²x+(a⁴-b⁴)=0

10. √3x²+11x+6√ 3=0

11. 4√ 3x²+5x-2√ 3=0

12. 3√7x²+4x-√7=0

13. √7x²-6x-13√7=0

14. 4√6x²-13x-2√6=0

15. x²-(1+√2)x+√2=0

16. 2x²+5√ 3x+6=0

17. x²-2x+1=0

18. 3x²+2√5x-5=0

19. 3a²x²+8abx+4b²=0,a≠ o

20. 2x²-2 √6x+3=0

21. 3x²-2x+2=0

22. √3x²+10x-8√ 3=0

23. x²+x+2=0

24. 16x²=24x+1

25. 25x²+20x+7=0

26. 6x²x-2=0

27. x²+5x+5=0

28. p²x²+(p²-q²)x-q²=0

29. abx²+(b²-ac)x-bc=0

30. x²-2ax+(a²-b²)=0

31. 12abx²-(9a²-8b²)x-6ab=0

32. 24x²– 41x + 12 = 0

33. 2x²– 7x – 15 = 0

34. 6x²+ 11x – 10 = 0

35. 10x² – 9x – 7 = 0

36. x² – x – 156 = 0

37. z²– 3²z – 105 = 0

38. 40 + 3x – x2 = 0

39. 6 – x – x² = 0

40. 7x²+ 49x + 84 = 0 

Question. A motor-boat goes 10 km upstream and returns back to the starting point in 55 minutes. If the speed of the motor boat in still water is 22 km/hr, find the speed of the current.
Answer: Let the speed of the current = \( x \) km/hr
\(\therefore\) The speed downstream = \( (22 + x) \) km/hr
The speed upstream = \( (22 - x) \) km/hr
Since, Time = \( \frac{\text{Distance}}{\text{Speed}} \)
\(\therefore\) Time for going 10 km downstream = \( \frac{10}{22+x} \) hrs
Time for returning back 10 km upstream = \( \frac{10}{22-x} \) hours
According to the condition,
\( \frac{10}{22+x} + \frac{10}{22-x} = \frac{55}{60} \)
\(\Rightarrow 10 \left[ \frac{1}{22+x} + \frac{1}{22-x} \right] = \frac{11}{12} \) [\(\because\) 55 minutes = \( \frac{55}{60} \) hours]
\(\Rightarrow 10 \times 12 \left[ \frac{22-x+22+x}{(22+x)(22-x)} \right] = 11 \)
\(\Rightarrow 120 \left[ \frac{44}{484-x^2} \right] = 11 \)
\(\Rightarrow 11(484 - x^2) = 120 \times 44 \)
\(\Rightarrow 5324 - 11x^2 = 5280 \)
\(\Rightarrow 11x^2 = 5324 - 5280 = 44 \)
\(\Rightarrow x^2 = \frac{44}{11} = 4 \)
\(\Rightarrow x = \pm 2 \)
But speed of the current cannot be negative,
\(\therefore x = 2 \)
\(\Rightarrow\) Speed of current = 2 km/hr

Question. A motor boat whose speed in still water is 5 km/hr, takes 1 hour more to go 12 km upstream than to return downstream to the same spot. Find the speed of the stream. 
Answer: Let the speed of the stream be \( x \) km/hr
\(\therefore\) Downstream speed of the motor boat = \( (x + 5) \) km/hr
\(\Rightarrow\) Time taken to go 12 km upstream = \( \frac{12}{5-x} \) hours
Time taken to return 12 km downstream = \( \frac{12}{5+x} \) hours
According to the condition
\( \frac{12}{5-x} - \frac{12}{5+x} = 1 \)
\(\therefore 12(5 + x) - 12(5 - x) = 1(5 - x)(5 + x) \)
\(\Rightarrow 60 + 12x - 60 + 12x = 25 - x^2 \)
\(\Rightarrow 24x = 25 - x^2 \)
\(\Rightarrow x^2 + 24x - 25 = 0 \)
\(\Rightarrow x^2 + 25x - x - 25 = 0 \)
\(\Rightarrow x(x + 25) - 1(x + 25) = 0 \)
\(\Rightarrow (x - 1)(x + 25) = 0 \)
Either \( x - 1 = 0 \Rightarrow x = 1 \)
or \( x + 25 = 0 \Rightarrow x = -25 \)
But \( x = -25 \) is not admissible, because the speed of the stream cannot be negative.
\(\therefore x = 1 \)
\(\Rightarrow\) speed of the stream = 1 km/hr.

Question. Sum of the areas of two squares is 260 m\(^2\). If the difference of their perimeters is 24 m, then find the sides of the two squares. 
Answer: Let the side of one of the squares be 'x' metres
\(\therefore\) Perimeter of square-I = \( 4 \times x \) metres = \( 4x \) metres
\(\therefore\) Perimeter of square-II = \( (24 + 4x) \) metres
\(\therefore\) Side of the square-II = \( \frac{1}{4}(24 + 4x) \) metres = \( (6 + x) \) metres
Now, according to the condition, we have:
\( x^2 + (6 + x)^2 = 260 \)
\(\Rightarrow x^2 + 36 + x^2 + 12x - 260 = 0 \)
\(\Rightarrow 2x^2 + 12x - 224 = 0 \)
\(\Rightarrow x^2 + 6x - 112 = 0 \) ...(1)
Comparing (1) with \( ax^2 + bx + c = 0 \), we get,
\( a = 1, b = 6, c = -112 \)
\(\therefore b^2 - 4ac = (6)^2 - 4(1)(-112) \)
\(= 36 + 448 = 484 \)
\(\therefore x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\(\Rightarrow x = \frac{-6 \pm \sqrt{484}}{2(1)} \)
\(= \frac{-6 \pm 22}{2} \)
Taking +ve sign, we have
\( x = \frac{-6 + 22}{2} = \frac{16}{2} = 8 \)
Taking -ve sign, we have
\( x = \frac{-6 - 22}{2} = \frac{-28}{2} = -14 \)
But \( x = -14 \) is not required, as the length of a side cannot be negative.
\(\therefore x = 8 \)
\(\Rightarrow\) Side of square-I = 8 m
\(\Rightarrow\) Side of square-II = \( 6 + 8 \) m = 14 m.

Question. The age of a father is twice the square of the age his son. Eight years hence, the age of the father will be 4 years more than three times the age of his son. Find their present ages. 
Answer: Let the present of son be 'x' years.
\(\therefore\) Father's present age = \( 2x^2 \) years
8 years hence:
Age of son = \( (x + 8) \) years
Age of father = \( (2x^2 + 8) \) years
According to the condition:
\( (2x^2 + 8) = 3(x + 8) + 4 \)
\(\Rightarrow 2x^2 + 8 - 3x - 24 - 4 = 0 \)
\(\Rightarrow 2x^2 - 3x + 8 - 28 = 0 \)
\(\Rightarrow 2x^2 - 3x - 20 = 0 \) ...(1)
Comparing (1) with \( ax^2 + bx + c = 0 \), we get
\( a = 2, b = -3, c = -20 \)
\(\therefore b^2 - 4ac = (-3)^2 - 4(2)(-20) \)
\(= 9 + 160 = 169 \)
Now, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\(\therefore x = \frac{-(-3) \pm \sqrt{169}}{2(2)} = \frac{3 \pm 13}{4} \)
Taking +ve sign,
\( x = \frac{3 + 13}{4} = \frac{16}{4} = 4 \)
Taking -ve sign,
\( x = \frac{3 - 13}{4} = \frac{-10}{4} = -\frac{5}{2} \)
But \( x = -\frac{5}{2} \) is not required, as the age cannot be negative.
\(\therefore x = 4 \)
\(\Rightarrow\) Present age of son = 4 years
Present age of father = \( 2 \times 4^2 = 32 \) years.

Question. A motor boat whose speed in still water is 16 km/h, takes 2 hours more to go 60 km upstream than to return to the same spot. Find the speed of the stream. 
Answer: Let the speed of the stream = \( x \) km/hr
For the motor boat, we have:
\(\therefore\) Downstream speed = \( (16 + x) \) km/hr
Upstream speed = \( (16 - x) \) km/hr
For going 60 km:
Downstream = \( \frac{60}{16+x} \) hours
Upstream = \( \frac{60}{16-x} \) hours
According to the condition:
\( \frac{60}{16-x} - \frac{60}{16+x} = 2 \)
\(\Rightarrow 60(16 + x) - 60(16 - x) = 2(16 - x)(16 + x) \)
\(\Rightarrow 960 + 60x - 960 + 60x = 2(256 - x^2) \)
\(\Rightarrow 2x^2 + 120x = 2 \times 256 - 2x^2 \)
\(\Rightarrow x^2 + 60x = 256 \)
\(\Rightarrow x^2 + 60x - 256 = 0 \) ...(1)
Comparing (1) with \( ax^2 + bx + c = 0 \),
\( a = 1, b = 60, c = -256 \)
\(\therefore b^2 - 4ac = (60)^2 - 4(1)(-256) \)
\(= 3600 + 1024 = 4624 \)
\(\therefore x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\(\Rightarrow x = \frac{-60 \pm \sqrt{4624}}{2(1)} \)
\(\Rightarrow x = \frac{-60 \pm 68}{2} \)
Taking +ve sign,
\( x = \frac{-60 + 68}{2} = \frac{8}{2} = 4 \)
Taking -ve sign,
\( x = \frac{-60 - 68}{2} = \frac{-128}{2} = -64 \)
Since, the speed of a stream cannot be negative,
\(\therefore x = -64 \) is not admissible
\(\therefore x = 4 \)
\(\Rightarrow\) speed of the stream = 4 km/hr.

Question. A train travels 288 km at a uniform speed. If the speed had been 4 km/hr more, it would have taken 1 hour less for the same journey. Find the speed of the train. 
Answer: Let the speed of the train be \( x \) km/hr
Total distance travelled = 288 km
\(\therefore\) Time taken = \( \frac{288}{x} \) hours
In the other case,
Speed of the train = \( (x + 4) \) km/hr
\(\therefore\) Time taken = \( \frac{288}{x+4} \) hours
According to the condition,
\( \frac{288}{x} - \frac{288}{x+4} = 1 \)
\(\Rightarrow \frac{288(x + 4) - 288x}{x(x + 4)} = 1 \)
\(\Rightarrow 288x + 1152 - 288x = 1(x)(x + 4) \)
\(\Rightarrow 288x + 1152 - 288x = x^2 + 4x \)
\(\Rightarrow 1152 = x^2 + 4x \)
\(\Rightarrow x^2 + 4x - 1152 = 0 \) ...(1)
Comparing (1) with \( ax^2 + bx + c = 0 \),
\( a = 1, b = 4, c = -1152 \)
\(\therefore b^2 - 4ac = (4)^2 - 4(1)(-1152) \)
\(= 16 + 4608 = 4624 \)
\(\Rightarrow x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\(\Rightarrow x = \frac{-(4) \pm \sqrt{4624}}{2(1)} = \frac{-4 \pm 68}{2} \)
Taking +ve sign,
\( x = \frac{-4 + 68}{2} = \frac{64}{2} = 32 \)
Taking -ve sign,
\( x = \frac{-4 - 68}{2} = \frac{-72}{2} = -36 \)
\(\because\) speed cannot be negative,
\(\therefore x \neq -36 \)
\(\therefore x = 32 \)
\(\Rightarrow\) speed of the train = 32 km/hr.

Question. A train travels at a certain average speed for a distance of 63 km and then travels a distance of 72 km at an average speed of 6 km/hr more than its original speed of. If takes 3 hours to complete the total journey, what is its original average speed? 
Answer: Let the original average speed = \( x \) km/hr
\(\therefore\) Time taken to cover 72 km = \( \frac{72}{x+6} \) hours
Time taken to cover 63 km = \( \frac{63}{x} \) hours
Since, total time = 3 hours
\(\therefore \frac{72}{x+6} + \frac{63}{x} = 3 \)
\(\Rightarrow \frac{1}{9} \left[ \frac{72}{x+6} + \frac{63}{x} \right] = \frac{1}{9} \times 3 \) [\(\because\) HCF of 72 and 63 is 9]
\(\Rightarrow \frac{8}{x+6} + \frac{7}{x} = \frac{1}{3} \)
\(\Rightarrow \frac{8x + 7(x + 6)}{x(x + 6)} = \frac{1}{3} \)
\(\Rightarrow 8x + 7x + 42 = \frac{x^2 + 6x}{3} \)
\(\Rightarrow 15x + 42 = \frac{x^2 + 6x}{3} \)
\(\Rightarrow 3[15x + 42] = x^2 + 6x \)
\(\Rightarrow 45x + 126 - x^2 - 6x = 0 \)
\(\Rightarrow x^2 - 39x - 126 = 0 \)
\(\Rightarrow x^2 - 42x + 3x - 126 = 0 \)
\(\Rightarrow x(x - 42) + 3(x - 42) = 0 \)
\(\Rightarrow (x + 3)(x - 42) = 0 \)
Either \( x + 3 = 0 \Rightarrow x = -3 \)
or \( x - 42 = 0 \Rightarrow x = 42 \)
Since, speed cannot be negative,
\(\therefore x = -3 \) is not desired.
Thus, the original speed of the train is 42 km/hr.

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, then find the values of p and k. (AI CBSE 2009)
Answer: Since -5 is a root of \( 2x^2 + px - 15 = 0 \),
\(\therefore\) Substituting \( x = -5 \) in the given equation, we get
\( 2(-5)^2 + p(-5) - 15 = 0 \)
\(\Rightarrow 2(25) + (-5p) - 15 = 0 \)
\(\Rightarrow 50 - 5p - 15 = 0 \)
\(\Rightarrow -5p + 35 = 0 \)
\(\Rightarrow -5p = -35 \)
\(\Rightarrow p = \frac{-35}{-5} = 7 \)
Now, comparing the another quadratic equation \( p(x^2 + x) + k = 0 \), i.e., \( px^2 + px + k = 0 \) with \( ax^2 + bx + c = 0 \), we have:
\( a = p \)
\( b = p \)
\( c = k \)
\( b^2 - 4ac = p^2 - 4(p)(k) = p^2 - 4pk \)
Since \( p(x^2 - x) + k = 0 \) has equal roots,
\(\therefore p^2 - 4pk = 0 \)
\(\Rightarrow (7)^2 - 4(7)k = 0 \) [\(\because p = 7\)]
\(\Rightarrow 49 - 28k = 0 \)
\(\Rightarrow k = \frac{-49}{-28} = \frac{7}{4} \)
Thus, the required values of \( p = 7 \) and \( k = \frac{7}{4} \).

Question. In a class test, the sum of Gagan’s marks in Mathematics and English is 45. If he had 1 more mark in Mathematics and 1 less in English, the product of marks would have been 500. Find the original marks obtained by Gagan in Mathematics and English separately. 
Answer: Let Gagan’s marks in maths \( = x \)
and Marks in English \( = (45 - x) \)
\( \therefore \) According to the condition,
\( (x + 1) \times (45 - x + 1) = 500 \)
\( \Rightarrow (x + 1) \times (44 - x) = 500 \)
\( \Rightarrow 44x - x^2 + 44 - x = 500 \)
\( \Rightarrow - x^2 + 44x - 456 - x = 0 \)
\( \Rightarrow x^2 - 43x + 456 = 0 \)
\( \Rightarrow x^2 - 19x - 24x + 456 = 0 \) [\( \because 24 \times 19 = 456 \), \( -43 = (-24) + (-19) \)]
\( \Rightarrow x (x - 19) - 24 (x - 19) = 0 \)
\( \Rightarrow (x - 19) (x - 24) = 0 \)
Either \( x - 19 = 0 \Rightarrow x = 19 \)
or \( x - 24 = 0 \Rightarrow x = 24 \)
When \( x = 19 \), then \( 45 - 19 = 26 \)
When \( x = 24 \), then \( 45 - 24 = 21 \)
\( \therefore \) Gagan’s marks in Maths \( = 19 \) and in English \( = 26 \)
Or
Gagan’s marks in Maths \( = 24 \) and in English \( = 21 \).

Question. The sum of areas of two squares is \( 640\text{ m}^2 \). If the difference of their perimeters is \( 64\text{ m} \), find the sides of two squares.
Answer: Let the side of square I be \( x \) metres.
\( \therefore \) Perimeter of square I \( = 4x \) metres
\( \Rightarrow \) Perimeter of square II \( = (64 + 4x)\text{ m} \)
\( \therefore \) Side of square II \( = \frac{1}{4} (64 + 4x)\text{ m} = (16 + x)\text{ m} \)
Now Area of square I \( = x \times x = x^2 \)
Area of square II \( = (16 + x) \times (16 + x) = (16 + x)^2 = 256 + x^2 + 32x \)
According to the condition,
Area of square I + Area of square II \( = 640 \)
\( \Rightarrow x^2 + [256 + x^2 + 32x] = 640 \)
\( \Rightarrow x^2 + x^2 + 32x + 256 - 640 = 0 \)
\( \Rightarrow 2x^2 + 32x - 384 = 0 \)
\( \Rightarrow x^2 + 16x - 192 = 0 \)
\( \Rightarrow x^2 + 24x - 8x - 192 = 0 \) [\( \because 24 - 8 = 16 \) and \( 24 \times 8 = 192 \)]
\( \Rightarrow x (x + 24) - 8 (x + 24) = 0 \)
\( \Rightarrow (x + 24) (x - 8) = 0 \)
Either \( x + 24 = 0 \Rightarrow x = -24 \)
or \( x - 8 = 0 \Rightarrow x = +8 \)
\( \because \) side of a square cannot be negative,
\( \therefore \) Rejecting \( x = -24 \), we have \( x = 8 \)
\( \Rightarrow \) Side of smaller square \( = 8\text{ m} \)
Side of larger square \( = 8 + 16\text{ m} = 24\text{ m} \).

Question. In a class test, the sum of Kamal’s marks in Mathematics and English is 40. Had he got 3 marks more in Mathematics and 4 marks less in English, the product of his marks would have been 360. Find his marks in two subjects separately. 
Answer: Let Kamal’s marks in Maths \( = x \)
\( \therefore \) His marks in English \( = (40 - x) \)
According to the condition,
\( (x + 3) [40 - x - 4] = 360 \)
\( \Rightarrow (x + 3) (36 - x) = 360 \)
\( \Rightarrow 36x - x^2 + 108 - 3x - 360 = 0 \)
\( \Rightarrow - x^2 + 33x - 252 = 0 \)
\( \Rightarrow x^2 - 33x + 252 = 0 \)
\( \Rightarrow x^2 - 21x - 12x + 252 = 0 \) [\( \because (-21) \times (-12) = 252 \), \( (-21) + (-12) = -33 \)]
\( \Rightarrow x (x - 21) - 12 (x - 21) = 0 \)
\( \Rightarrow (x - 21) (x - 12) = 0 \)
Either \( (x - 21) = 0 \Rightarrow x = 21 \)
or \( (x - 12) = 0 \Rightarrow x = 12 \)
For \( x = 21 \), Marks of Kamal in Maths \( = 21 \), in English \( = 40 - 21 = 19 \)
For \( x = 12 \), Marks of Kamal in Maths \( = 12 \), in English \( = 40 - 12 = 28 \).

Question. Find the value of \( p \) for which the quadratic equation \( 4x^2 + px + 3 = 0 \) has equal roots. 
Answer: For equal roots, \( b^2 - 4ac = 0 \)
\( \therefore p^2 - 4(4)(3) = 0 \) or \( p^2 - 48 = 0 \Rightarrow p = \pm \sqrt{48} = \pm 4\sqrt{3} \)

Question. Solve the quadratic equation \( 2x^2 + ax - a^2 = 0 \). 
Answer: \( 2x^2 + ax - a^2 = 0 \)
\( \Rightarrow 2x^2 + 2ax - ax - a^2 = 0 \) [splitting '\( ax \)' into '\( 2ax \)' and '\( -ax \)']
\( \Rightarrow 2x[x + a] - a[x + a] = 0 \)
\( \Rightarrow (x + a) (2x - a) = 0 \)
\( x = -a \) or \( x = \frac{a}{2} \)

HOTS QUESTIONS

Question. Had Ravita scored 10 more marks in her Mathematics test out of 30 marks, 9 times these marks would have been the square of her actual marks. How many marks did she get in the test? 
Answer: Let actual marks be \( x \)
\( \therefore 9 \times [ \text{Actual marks} + 10] = [\text{Square of actual marks}] \)
or \( 9 \times (x + 10) = x^2 \)
\( \Rightarrow 9x + 90 = x^2 \)
\( \Rightarrow x^2 - 9x - 90 = 0 \)
\( \Rightarrow x^2 - 15x + 6x - 90 = 0 \)
\( \Rightarrow x(x - 15) + 6(x - 15) = 0 \)
\( \Rightarrow (x + 6) (x - 15) = 0 \)
Either \( x + 6 = 0 \Rightarrow x = -6 \)
or \( x - 15 = 0 \Rightarrow x = 15 \)
But marks cannot be less than 0.
\( \therefore x = -6 \) is not desired.
Thus, Ravita got 15 marks in her Mathematics test.

Question. A motor boat whose speed is \( 18\text{ km/h} \) in still water takes 1 hour more to go \( 24\text{ km} \) upstream than to return downstream to the same spot. Find the speed of the stream. 
Answer: Let the speed of the stream \( = x\text{ km/hr} \)
\( \therefore \) speed of the motor boat:
upstream \( = (18 - x)\text{ km/hr} \)
downstream \( = (18 + x)\text{ km/hr} \)
\( \Rightarrow \) Time taken by the motor boat in going:
\( 24\text{ km} \) downstream \( = \frac{24}{18 + x} \) hours [\( \because \text{Time} = \frac{\text{Distance}}{\text{Speed}} \)]
\( 24\text{ km} \) upstream \( = \frac{24}{18 - x} \) hours
According to the condition:
\( \frac{24}{18 - x} - \frac{24}{18 + x} = 1 \)
\( \Rightarrow 24 \times (18 + x) - 24 (18 - x) = 1 (18 - x) (18 + x) \)
\( \Rightarrow 24 [18 + x - 18 + x] = 18^2 - x^2 \)
\( \Rightarrow 24 [2x] = 324 - x^2 \)
\( \Rightarrow 48x = 324 - x^2 \)
\( \Rightarrow x^2 + 48x - 324 = 0 \)
\( \Rightarrow x^2 + 54x - 6x - 324 = 0 \)
\( \Rightarrow x (x + 54) - 6 (x + 54) = 0 \)
\( \Rightarrow (x - 6) (x + 54) = 0 \)
Either \( x - 6 = 0 \Rightarrow x = 6 \)
or \( x + 54 = 0 \Rightarrow x = -54 \)
But speed cannot be negative
\( \therefore \) Rejecting \( x = -54 \), we have \( x = 6 \Rightarrow \) Speed of the boat \( = 6\text{ km/hr} \).

Question. In a class test, the sum of marks obtained by P in Mathematics and Science is 28. Had he got 3 more marks in Mathematics and 4 marks less in Science, the product of marks obtained in the two subjects would have been 180? Find the marks obtained in two subjects separately. 
Answer: Let marks obtained by P in Maths be ‘\( x \)’.
\( \therefore \) His marks in Science \( = (28 - x) \)
According to the condition,
\( (x + 3) (28 - x - 4) = 180 \)
\( \Rightarrow (x + 3) (- x + 24) = 180 \)
\( \Rightarrow 24x - x^2 + 72 - 3x = 180 \)
\( \Rightarrow - x^2 + 21x + 72 - 180 = 0 \)
\( \Rightarrow - x^2 + 21x - 108 = 0 \)
\( \Rightarrow x^2 - 21x + 108 = 0 \)
\( \Rightarrow x^2 - 12x - 9x + 108 = 0 \)
\( \Rightarrow x (x - 12) - 9(x - 12) = 0 \)
\( \Rightarrow (x - 9) (x - 12) = 0 \)
Either \( x - 9 = 0 \Rightarrow x = 9 \)
or \( x - 12 = 0 \Rightarrow x = 12 \)
When \( x = 9 \) then \( 28 - x = 28 - 9 = 19 \)
When \( x = 12 \) then \( 28 - x = 28 - 12 = 16 \)
Thus P’s marks in Maths \( = 9 \) and Science \( = 19 \)
Or
P’s marks in Maths \( = 12 \) and Science \( = 16 \).

Question. Solve for \( x \): \( \sqrt{3}x^2 - 2\sqrt{2}x - 2\sqrt{3} = 0 \). 
Answer: \( \sqrt{3}x^2 - 2\sqrt{2}x - 2\sqrt{3} = 0 \), we have \( a = \sqrt{3} \), \( b = -2\sqrt{2} \) and \( c = -2\sqrt{3} \)
\( \therefore b^2 - 4ac = (-2\sqrt{2})^2 - 4(\sqrt{3})(-2\sqrt{3}) = 8 + 24 = 32 \)
Using Quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
we get, \( x = \sqrt{6} \)

Question. Solve for \( x \): \( x^2 + 5\sqrt{5}x - 70 = 0 \).
Answer: \( a = 1 \), \( b = 5\sqrt{5} \) and \( c = -70 \)
\( \Rightarrow b^2 - 4ac = (5\sqrt{5})^2 - 4(1)(-70) = 125 + 280 = 405 = 81 \times 5 \)
Now use quadratic formula to get \( x = 2\sqrt{5} \), \( -7\sqrt{5} \)

Question. At ‘\( t \)’ minutes past \( 2\text{ pm} \), the time needed by the minute hand of a clock to show \( 3\text{ pm} \) was found to be 3 minutes less than \( \frac{t^2}{4} \) minutes. Find ‘\( t \)’.
Answer: For a minute-hand time needed to show \( 2\text{ pm} \) to \( 3\text{ pm} \) is ‘\( 60 \)’ minutes.
It has already covered ‘\( t \)’ minutes.
\( \therefore \) Time required by the minute-hand to reach to 12 (at \( 3\text{ pm} \)) \( = (60 - t) \) minutes.
\( \therefore \left( \frac{t^2}{4} - 3 \right) = (60 - t) \)
\( \Rightarrow \frac{t^2}{4} + t - 63 = 0 \)
\( \Rightarrow t^2 + 4t - 252 = 0 \)
Solving, we get, \( t = 14 \) or \( -18 \)
But \( t = -18 \) is not desirable (being negative)
Thus, \( t = 14 \) minutes.

Question. A train, travelling at a uniform speed for \( 360\text{ km} \), would have taken 48 minutes less to travel the same distance if its speed were \( 5\text{ km/hr} \) more. Find the original speed of the train. 
Answer: Let the original speed be \( x\text{ km/hr} \)
\( \therefore \) Original time taken \( = \frac{360}{x} \) hours
New speed \( = (x + 5)\text{ km/hr} \)
\( \therefore \) New time \( = \frac{360}{x + 5} \) hours
According to the condition,
\( \frac{360}{x} = \frac{360}{x + 5} + \frac{48}{60} \)
\( \Rightarrow 360 \left[ \frac{1}{x} - \frac{1}{x + 5} \right] = \frac{4}{5} \)
\( \Rightarrow 360 \left[ \frac{x + 5 - x}{x(x + 5)} \right] = \frac{4}{5} \)
\( \Rightarrow \frac{360 \times 5}{x^2 + 5x} = \frac{4}{5} \Rightarrow x^2 + 5x - 2250 = 0 \)
Solving for \( x \), we get \( x = -50 \) or \( 45 \)
Speed cannot be negative
\( \therefore \) Rejecting \( x = -50 \), we have \( x = 45 \)
Thus, the original speed of the train \( = 45\text{ km/hr} \).

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: For equal roots \( D = 0 \)
Here, \( D = (c - a)^2 - 4(b - c)(a - b) \)
\( \therefore (c - a)^2 - 4(b - c)(a - b) = 0 \)
\( \Rightarrow c^2 + a^2 + 4b^2 - 2ac - 4ab + 4ac - 4bc = 0 \)
\( \Rightarrow (c + a - 2b)^2 = 0 \)
\( \Rightarrow (c + a - 2b) = 0 \Rightarrow c + a = 2b \) (Hence Proved)

Question. If the roots of the equations \( ax^2 + 2bx + c = 0 \) and \( bx^2 - 2\sqrt{ac}x + b = 0 \) are simultaneously real then prove that \( b^2 = ac \).
Answer: Let \( D_1 \) and \( D_2 \) be the discriminants.
\( \therefore D_1 \geq 0 \) and \( D_2 \geq 0 \)
\( \Rightarrow 4b^2 - 4ac \geq 0 \) and \( 4ac - 4b^2 \geq 0 \)
\( \Rightarrow b^2 \geq ac \) and \( ac \geq b^2 \)
\( b^2 = ac \)

Question. If the roots of the equation \( (c^2 - ab)x^2 - 2(a^2 - bc)x + b^2 - ac = 0 \) are equal, then prove that either \( a = 0 \) or \( a^3 + b^3 + c^3 = 3abc \).
Answer: \( D = B^2 - 4AC \)
\( = [-2(a^2 - bc)]^2 - 4 (c^2 - ab) (b^2 - ac) \)
\( = 4a (a^3 + b^3 + c^3 - 3abc) \)
For equal roots, \( D = 0 \)
\( \therefore 4a (a^3 + b^3 + c^3 - 3abc) = 0 \)
\( \Rightarrow a = 0 \) or \( a^3 + b^3 + c^3 = 3abc \)

Please click the link below to download CBSE Class 10 Quadratic Equations Sure Shot Questions Set C.