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SEBA Class 10 Mathematics Chapter 4 Quadratic Equation
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Quadratic Equation
Chapter – 4
| Exercise 4.1 |
1. Check whether the following are quadratic equations:
(i) (x + 1)2 = 2( x-3)
Ans: (i) (x + 1)2 = 2( x-3)
⇒ x2 + 2x +1= 2x – 6 ⇒ x2 +7 = 0
⇒ x2 + 0. X + 7 = 0
It is of the from ax2 + bx + c = 0
The given equation is a quadratic equation.
(ii) (x-2) (x-3) =(x-1) ( x +3)
Ans: x²-2x=(-2)(3-x)
⇒ x²-2x=-6+2x
⇒ x²- 4x+6= 0
It is of the form ax² + bx + c = 0
∴ The given equation is quadratic
(iii) x -2) (x-1) = (x-1) = (x-1) (x + 3)
Ans: (x-2) (x+1) = (x-1) (x+3)
⇒ x²+x-2x-2=x²+3x-x-3
⇒ x-x-2=x²+2x-2x-3
⇒ -x-2x-2+3=0
⇒ -3x+1=0 or 3x-1=0
It is not of the form ax² + bx + c = 0
∴ The given equation is not quadratic equation.
(iv) (x – 3) (2x +1) = x (x+5)
Ans: (x-3) (2x+1)=x(x+5)
2x² + x-6x – 3 = x² + 5x
→2-5x-3 = x² + 5x
2 -x- 5x – 5x – 2 = 0
⇒ -10x-3=0 It is a quadratic equation.
(v) (2x – 1) (x-3) = (x+5) (x-1)
Ans:(x-3) (2x+1)=x(x+5)
2x² + x-6x – 3 = x² + 5x
→ 2 – 5x -3 = x² + 5x
2 -x -5x – 5x – 2 = 0
⇒ -10x-3 = 0 It is a quadratic equation.
(vi) x2 +3x + 1 = (x-)2
Ans: x²+3 x +1=(x-2)
X² + 3x + 1= x² – 4x + 4
3x + 4x + 1-4=0
⇒7x-3=0 It is a linear equation and not a quadratic equation.
(vii) (x-3)3 = 2x (x2 -1 )
Ans: (x+2)² = 2x (x² – 1)
⇒x² + 3x² x 2 + 3x x 2² + 23 = 2x² – 2x
⇒ x + 6x² + 12x + 8 = 2x²-2x
⇒-x + 6x² + 14x + 8 = 0
⇒x²-6x² – 14x – 8 = 0
It is a cubic equation and not a : quadratic equation.
(viii) x3 – 4 x2 – x +1 = (x- 2)3
Ans: x²-4x²-x+1=(x-2)
⇒ -4x² – x + 1 = x² + 3(x)(-2)(x-2)-2
⇒ x² – 4x² – x + 1 = x² – 6x(x-2)-8
⇒ x-4x²-x + 1 = x²-6x² + 12x-8
→ 4x² + 6x² – x – 2x + 1 + 8 = 02-13x + 9 = 0 It is a quadratic equation.
2. Represent the following situations in the form of quadratic equation:
(i) The area of a rectangular plot is 528 m². The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
Ans: Let breadth be = x metres
Then length =(2x+1) metres
X x (2x+1) = 528 (Area of the plot)
Or 2x² + x – 528 = 0
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
Ans: Let the positive consecutive integers be x and x + 1 We are given that x x (x + 1) = 306 x² + x-306 = 0
(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from will be 360. We would like to find Rohan’s present age.
Ans: Rohan’s present age = x years Then present age of Rohan’s mother = (x + 26) years.
After three years, its is given that,
( x+3) x [(x+26 ) + 3] = 360
(x+3) x (x+29 ) = 360
X² + 32 x+87 = 360
Or
x² + 32x + 87-360 = 0 x² + 32x – 273 = 0
(iv) A train travels a distance of 480 km. at a uniform speed. If the speed had been 8km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train. Solution: (i) Let breadth be = x metres
Ans: Do yourself.
| Exercise 4.2 |
1. Find the roots of the following quadratic equations Factorisation:
(i) x²-3x-10 = 0
Ans: (i) x²-3x-100x²-5x+2x-10=0
⇒ x(x-5)+2(x – 5) = 0
⇒ (x-5) (x + 2) = 0
⇒ (x-5) = 0 or (x + 2) = 0
⇒ x = 5 or x=-2
Hence, the two roots are -2 and 5
(ii) 2x2 + x – 6 = 0
Ans: 2x²+x-6=02x²+4x-3x-6=0
2x (x+2)-3 (x+2) = 0
(x+2) (2x-3)=0
⇒ X+2 = 0 or 2x-3 = 0
⇒ x = -2 or x = 2
(iii) √2x² + 7x +5√2=0
Ans: √2x² + 7x 5√2=0
√2x² +5+2x +5√2 = 0
x(√2x +5 + √2(√2 +5) = 0
(x + √2x) (√2 +5) =0
(iv) 2x2 – x + 1/8 = 0
Ans: Do yourself
(v) 100x²-20x + 1 = 0
Ans: Do yourself
2. Represent the following situations mathematically:
(i) John Jivanti together has 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 324. We would like to find out how many marbles they had to start with.
Ans: The quadratic equation is * x ^ 2 – 45x + 324 = 0 Where x is the number of marbles with John. The equation can be rewritten as
X² – 36x – 9x + 324 = 0
⇒x(x-36)-9(x-36) = 0
⇒ (x-9) (x−36) = 0 ⇒ x = 9 or 36
Therefore , either john had 9 marbles and jivanti had 36 marbles or vice versa
(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was Rs. 750. We would like to find out the number of toys produced on that day.
3. Find the numbers whose sum is 27 and product is 182
4. Find two consecutive positive integers, sum of whose squares is 365.
5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm., find the other two sides.
Ans: In ΔΑΒC, base BC and altitude AC = x cm. = (x-7) cm. ∠ACB = 90°, AB = 13cm.
By Pythagoras Theorem, we have, BC2 + AC2 = AB2 ⇒ x² + (x – 7)² = 132
⇒x²+x² – 14x + 49 = 169
2x²-14x- 120 = 0
⇒x²-7x-60 = 0
⇒x² – 12x + 5x-60 = 0
⇒x(x-12) +5(x – 12) = 0
⇒1 – 5x – 12
Sides are not negative
12 2.BC = 12 cm and AC = 5cm.
6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs. 90, find the number of articles produced and the cost of each article.
Ans: Let the number of articles produced in a day (on a particular day )= x
x(2x + 3) = 90
We are given that,
⇒ 2x² + 3x – 90 = 0
⇒ 2x² + 15x-12x – 90 = 0
⇒x(2x+15) – 6(2x + 15) = 0
⇒ (2x+15) (x-6) = 0
Now v, 2x + 15 =0 Rightarrow 2x=-15 Or, x -6=0 Rightarrow x = 6 15 x=-2 (not exist)
Hence, the number of articles = 6
Cost of production of one article = Rs.(2 x 6 + 3) = Rs. 15
| Exercise 4.3 |
1. Find the roots of the following quadratic equations, if they exist, by the method of completing square :
(i) 2x²-7x + 3 = 0
Ans:
(ii) 2x² + x – 4 = 0
Ans:
(iii) 4x² + 4√3x +3=0
Ans:
(iv) 2x² + x + 4 = 0
Ans:
2. Find the roots of the quadratic equations given Q.1 above by applying the quadratic formula.
(i) 2x² – 7x + 3 = 0
Here a = 2, b = -7, c = 3
Discriminant D = (-7)² – 4(2) (3) = 49-24 = 25
⇒ √D = √25=5
(ii) 2x² + x − 4 = 0
Ans: 2x² + x − 4 = 0
Here a = 2, b = 1, c = -4
D = (1)² – 4(2)(-4) = 1 + 32 = 33
⇒√D = √33
3.Find the roots of the following equations:
Ans:
Ans:
4. The sum of the reciprocal of Rehman’s ages (in years) 3 years ago
Ans:
5. 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 their marks would have been 210. Find her marks in two subjects.
Ans: Let the marks of Shefali in Mathematics be x Then marks in English will be 30 – x We are given that,
(x+2) x (30-x-3) = 210 (x + 2) (27 – x) = 210
⇒x² + 25x + 54 = 210
⇒ x² – 25x + 210-54 = 0
x²-25x + 156 = 0
Here, a = 1, b = -25, c = 156
D = (-25)²-4 x 1 x 156 = 625-624 =1
√D = 1
When x = 12, we have marks in Mathematics s=12 and marks in English
= 18 When x = 13, we have marks in Mathematics = 13 and marks in English = 17
6. 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.
Ans:
7. The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
Ans:
8. A train travels 360 km at a uniform speed. If the speed had been 5km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Ans:
Ans:
11. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 1 km/h more than that of the passenger train, find the average speed of the two trains.
Ans: Let the side of the smaller square be x m.
Then, perimetre of the smaller square = 4x m.
Perimetre of the larger square = (4x +24)m Therefore,
| Exercise 4.4 |
1. Find the nature of the roots of the following quadratic equations. If real roots exist, find them.
(i) 2x² – 3x + 5 = 0
Ans: 2x² – 3x + 5 = 0
a = 2, b = -3, c = 5
Discriminant D = b² – 4ac = 9 – 4x2x5=9-40-31
⇒ D< 0 Hence, no real root.
(ii) 3x² – 4/3x + 4 = 0
Ans:
(iii) 2x2 – 6x + 3 = 0
Ans: 2x²-6x + 3 = 0
a = 2, b = -6, c = 3
Discriminant D = b²-4ac = 36 – 24 = 12
⇒ D > 0
2. Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) 2x² + kx + 3 = 0
Ans: 2x2 + k x + 3 = 0
Comparing equation with ax2 + bx + c = 0,
we obtain a = 2, b = k, c = 3
Discriminant = b2 − 4ac = (k)2− 4(2) (3) = k2 − 24
For equal roots,
Discriminant = 0
k2 − 24 = 0
k2 = 24
(ii) k x (x − 2) + 6 = 0
or kx2 − 2kx + 6 = 0
Comparing this equation with ax2 + bx + c = 0,
we obtain a = k, b = −2k, c = 6
Discriminant = b2 − 4ac = (− 2k)2 − 4 (k) (6) = 4k2 − 24k
For equal roots, b2 − 4ac = 0
4k2 − 24k = 0
4k (k − 6) = 0
Either 4k = 0 or k = 6 = 0
k = 0 or k = 6
However, if k = 0, then the equation will not have the terms ‘x2’ and ‘x’.
Therefore, if this equation has two equal roots, k should be 6 only.
3.Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2?If so, find its length and breadth.
Ans: Let the breadth of the rectangular mango grove hem. Then the length of the rectangular mango grove = 2x m.
Therefore, Area of the rectangular mango grove = length x breadth = 2x x x m² = 2x² m² According to question, 2x² = 800x² = 400 = (20)⇒ x = 20 Hence, the rectangle is possible and it has breadth = 20m and length = 40m.
4. Is the following situation possible? If so, determine their ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
Ans: (x-4)x(16x) = 48 ⇒16x-x²-64 + 4x = 48
⇒-x² + 20x – 64 – 48 = 0
⇒ x² + 20x – 112 = 0
⇒x²-20x + 112 = 0
a = 1, b = -20, c = 112
Discriminant D = b² – 4ac = 400 – 448 = -48 i.e., D < 0
Thus, no real value of x.
Hence, the situation is not possible.
5. Is it possible to design a rectangular park of perimetre 80 m and area 400m³? If so, find its length and breadth.
Ans: Perimetre of the rectangular park = 80 m.
Length + Breadth of the park 2 m = 40 m.
Let the breadth be x metres, then length = (40 – x)m.
Here, x < 40
x2 (40-x) = 400 -x² + 40x – 400 = 0
⇒ x²-40x + 400 = 0
(x – 20)² = 0 ⇒ x = 20m
Thus, we have a length of breadth = 20m.
Therefore, the park is a square having a 20m side.
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chapter 4.3