Quadratic Equations Ex 4.1 Solutions
Question - 1 : - Choose the correct answer from the given fouroptions in the following questions:
Answer - 1 : -
1. Which of the following is a quadraticequation?
(A) x2 + 2x +1 = (4 – x)2 + 3 (B) –2x2 =(5 – x)(2x-(2/5))
(C) (k + 1) x2 +(3/2) x = 7, where k = –1 (D) x3 – x2 =(x – 1)3
Solution:
(D) x3 – x2 =(x – 1)3
Explanation:
The standard form of a quadratic equation isgiven by,
ax2 + bx + c = 0, a ≠ 0
(A) Given, x2 + 2x + 1 = (4 –x)2 + 3
x2 + 2x + 1 = 16 – 8x + x2 +3
10x – 18 = 0
which is not a quadratic equation.
(B) Given, -2x2 = (5 – x) (2x– 2/5)
-2x2 = 10x – 2x2 –2 +2/5x
52x – 10 = 0
which is not a quadratic equation.
(C) Given, (k + 1) x2 + 3/2 x = 7, where k = -1
(-1 + 1) x2 + 3/2 x = 7
3x – 14 = 0
which is not a quadratic equation.
(D) Given, x3 – x2 =(x – 1)3
x3 – x2 = x3 –3x2 + 3x – 1
2x2 – 3x + 1 = 0
which represents a quadratic equation.
Question - 2 : - Choose the correct answer from the given fouroptions in the following questions:-
Answer - 2 : -
Which of the following is not a quadraticequation?
(A) 2(x – 1)2 = 4x2 –2x + 1 (B) 2x – x2 = x2 +5
(C) (√2x + √3)2 + x2 =3x2 − 5x (D) (x2 + 2x)2 = x4 +3 + 4x3
Solution:
(D) (x2 + 2x)2 = x4 +3 + 4x3
A quadratic equation is represented by theform,
ax2 + bx + c = 0, a ≠ 0
(A) Given, 2(x – 1)2 = 4x2 –2x + 1
2(x2 – 2x + 1) = 4x2 –2x + 1
2x2 + 2x – 1 = 0
which is a quadratic equation.
(B) Given, 2x – x2 = x2 +5
2x2 – 2x + 5 = 0
which is a quadratic equation.
(C) Given, (√2x + √3)2 =3x2 – 5x
2x2 + 2√6x + 3 = 3x2 –5x
x2 – (5 + 2√6)x – 3 = 0
which is a quadratic equation.
(D) Given, (x2 + 2x)2 =x4 + 3 + 4x2
x4 + 4x3 + 4x2 =x4 + 3 + 4x2
4x3 – 3 = 0
which is a cubic equation and not a quadraticequation.
Question - 3 : - Choose the correct answer from the given fouroptions in the following questions--
Answer - 3 : -
If ½ is a root of the equation x2 +kx – 5/4 = 0, then the value of k is
(A) 2 (B) – 2
(C) ¼ (D) ½
Solution:
(A) 2
If ½ is a root of the equation
x2 + kx – 5/4 = 0 then,substituting the value of ½ in place of x should give us the value of k.
Given, x2 + kx – 5/4 = 0where, x = ½
(½)2 + k (½) – (5/4) = 0
(k/2) = (5/4) – ¼
k = 2
Question - 4 : - Choose the correct answer from the given four options in the following questions-->
Answer - 4 : -
Which of the following equations has the sumof its roots as 3?
(A) 2x2 – 3x + 6 = 0 (B) –x2 +3x – 3 = 0
(C) √2x2 – 3/√2x+1=0 (D) 3x2 –3x + 3 = 0
Solution:
(B) –x2 + 3x – 3 = 0
The sum of the roots of a quadratic equationax2 + bx + c = 0, a ≠ 0 is given by,
Coefficient of x / coefficient of x2 =– (b/a)
(A) Given, 2x2 – 3x + 6 = 0
Sum of the roots = – b/a = -(-3/2) = 3/2
(B) Given, -x2 + 3x – 3 = 0
Sum of the roots = – b/a = -(3/-1) = 3
(C) Given, √2x2 –3/√2x+1=0
2x2 – 3x + √2 = 0
Sum of the roots = – b/a = -(-3/2) = 3/2
(D) Given, 3x2 – 3x + 3 = 0
Sum of the roots = – b/a = -(-3/3) = 1
Question - 5 : - Choose the correct answer from the given fouroptions in the following questions.
Answer - 5 : -
Which of the following equations has 2 as aroot?
(A) x2 – 4x +5 = 0 (B) x2 + 3x – 12 = 0
(C) 2x2 – 7x +6 = 0 (D) 3x2 – 6x – 2 = 0
Solution:
(C) 2x2 – 7x +6 = 0
If 2 is a root then substituting the value 2in place of x should satisfy the equation.
(A) Given,
x2 – 4x + 5 = 0
(2)2 – 4(2) + 5 = 1 ≠ 0
So, x = 2 is not a root of x2 –4x + 5 = 0
(B) Given, x2 + 3x – 12 = 0
(2)2 + 3(2) – 12 = -2 ≠ 0
So, x = 2 is not a root of x2 +3x – 12 = 0
(C) Given, 2x2 – 7x + 6 = 0
2(2)2 – 7(2) + 6 = 0
Here, x = 2 is a root of 2x2 –7x + 6 = 0
(D) Given, 3x2 – 6x – 2 = 0
3(2)2 – 6(2) – 2 = -2 ≠ 0
So, x = 2 is not a root of 3x2 –6x – 2 = 0
Question - 6 : - Check whether the following are quadratic equations:
Answer - 6 : -
(i) (x + 1)2 = 2(x – 3)
(ii) x2 – 2x = (–2) (3 – x)
(iii) (x – 2)(x + 1) = (x – 1)(x + 3)
(iv) (x – 3)(2x +1) = x(x + 5)
(v) (2x – 1)(x – 3) = (x + 5)(x – 1)
(vi) x2 + 3x + 1 = (x – 2)2
(vii) (x + 2)3 = 2x (x2 – 1)
(viii) x3 – 4x2 – x + 1 = (x – 2)3
Solutions:
(i) Given,
(x + 1)2 = 2(x – 3)
By using the formula for (a+b)2 = a2+2ab+b2
⇒ x2 + 2x + 1 = 2x – 6
⇒ x2 + 7 = 0
Since the above equation is in the form of ax2 + bx + c = 0.
Therefore, the given equation is quadratic equation.
(ii) Given, x2 – 2x = (–2) (3 – x)
By using the formula for (a+b)2 = a2+2ab+b2
⇒ x2 – 2x = -6 + 2x
⇒ x2 – 4x + 6 = 0
Since the above equation is in the form of ax2 + bx + c = 0.
Therefore, the given equation is quadratic equation.
(iii) Given, (x – 2)(x + 1) = (x – 1)(x + 3)
By using the formula for (a+b)2 = a2+2ab+b2
⇒ x2 – x – 2 = x2 + 2x – 3
⇒ 3x – 1 = 0
Since the above equation is not in the form of ax2 + bx + c = 0.
Therefore, the given equation is not a quadratic equation.
(iv) Given, (x – 3)(2x +1) = x(x + 5)
By using the formula for (a+b)2=a2+2ab+b2
⇒ 2x2 – 5x – 3 = x2 + 5x
⇒ x2 – 10x – 3 = 0
Since the above equation is in the form of ax2 + bx + c = 0.
Therefore, the given equation is quadratic equation.
(v) Given, (2x – 1)(x – 3) = (x + 5)(x – 1)
By using the formula for (a+b)2=a2+2ab+b2
⇒ 2x2 – 7x + 3 = x2 + 4x – 5
⇒ x2 – 11x + 8 = 0
Since the above equation is in the form of ax2 + bx + c = 0.
Therefore, the given equation is quadratic equation.
(vi) Given, x2 + 3x + 1 = (x – 2)2
By using the formula for (a+b)2=a2+2ab+b2
⇒ x2 + 3x + 1 = x2 + 4 – 4x
⇒ 7x – 3 = 0
Since the above equation is not in the form of ax2 + bx + c = 0.
Therefore, the given equation is not a quadratic equation.
(vii) Given, (x + 2)3 = 2x(x2 – 1)
By using the formula for (a+b)2 = a2+2ab+b2
⇒ x3 + 8 + x2 + 12x = 2x3 – 2x
⇒ x3 + 14x – 6x2 – 8 = 0
Since the above equation is not in the form of ax2 + bx + c = 0.
Therefore, the given equation is not a quadratic equation.
(viii) Given, x3 – 4x2 – x + 1 = (x – 2)3
By using the formula for (a+b)2 = a2+2ab+b2
⇒ x3 – 4x2 – x + 1 = x3 – 8 – 6x2 + 12x
⇒ 2x2 – 13x + 9 = 0
Since the above equation is in the form of ax2 + bx + c = 0.
Therefore, the given equation is quadratic equation.
Question - 7 : - Represent the following situations in the form of quadratic equations:
Answer - 7 : -
(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.
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken
Solutions:
(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.
Solutions:
(i) Let us consider,
Breadth of the rectangular plot = x m
Thus, the length of the plot = (2x + 1) m.
As we know,
Area of rectangle = length × breadth = 528 m2
Putting the value of length and breadth of the plot in the formula, we get,
(2x + 1) × x = 528
⇒ 2x2 + x =528
⇒ 2x2 + x – 528 = 0
Therefore, the length and breadth of plot, satisfies the quadratic equation, 2x2 + x – 528 = 0, which is the required representation of the problem mathematically.
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
Solutions:
(ii) Let us consider,
The first integer number = x
Thus, the next consecutive positive integer will be = x + 1
Product of two consecutive integers = x × (x +1) = 306
⇒ x2 + x = 306
⇒ x2 + x – 306 = 0
Therefore, the two integers x and x+1, satisfies the quadratic equation, x2 + x – 306 = 0, which is the required representation of the problem mathematically.
(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
Solutions:
(iii) Let us consider,
Age of Rohan’s = x years
Therefore, as per the given question,
Rohan’s mother’s age = x + 26
After 3 years,
Age of Rohan’s = x + 3
Age of Rohan’s mother will be = x + 26 + 3 = x + 29
The product of their ages after 3 years will be equal to 360, such that
(x + 3)(x + 29) = 360
⇒ x2 + 29x + 3x + 87 = 360
⇒ x2 + 32x + 87 – 360 = 0
⇒ x2 + 32x – 273 = 0
Therefore, the age of Rohan and his mother, satisfies the quadratic equation, x2 + 32x – 273 = 0, which is the required representation of the problem mathematically.
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken
Solutions:
(iv) Let us consider,
The speed of train = x km/h
And
Time taken to travel 480 km = 480/x km/hr
As per second condition, the speed of train = (x – 8) km/h
Also given, the train will take 3 hours to cover the same distance.
Therefore, time taken to travel 480 km = 480/(x+3) km/h
As we know,
Speed × Time = Distance
Therefore,
(x – 8)(480/(x + 3) = 480
⇒ 480 + 3x – 3840/x – 24 = 480
⇒ 3x – 3840/x = 24
⇒ 3x2 – 8x – 1280 = 0
Therefore, the speed of the train, satisfies the quadratic equation, 3x2 – 8x – 1280 = 0, which is the required representation of the problem mathematically.
Question - 8 : - Check whether the following are quadratic equations:-
Answer - 8 : -
(i) (x + 1)2 =2(x – 3)
(ii) x2 –2x = (–2) (3 – x)
(iii) (x – 2)(x + 1)= (x – 1)(x + 3)
(iv) (x – 3)(2x +1)= x(x + 5)
(v) (2x – 1)(x – 3)= (x + 5)(x – 1)
(vi) x2 +3x + 1 = (x – 2)2
(vii) (x + 2)3 =2x (x2 – 1)
(viii) x3 –4x2 – x + 1 = (x – 2)3
Solutions:
(i) Given,
(x + 1)2 = 2(x – 3)
By using the formula for (a+b)2 =a2+2ab+b2
⇒ x2 + 2x + 1 = 2x – 6
⇒ x2 + 7 = 0
Since the above equation is in the form of ax2 + bx + c =0.
Therefore, the given equationis quadratic equation.
(ii) Given, x2 – 2x = (–2) (3– x)
By using the formula for (a+b)2 =a2+2ab+b2
⇒ x2 – 2x = -6 + 2x
⇒ x2 – 4x + 6 = 0
Since the above equation is in the form of ax2 + bx + c =0.
Therefore, the given equationis quadratic equation.
(iii) Given, (x – 2)(x + 1) = (x – 1)(x + 3)
By using the formula for (a+b)2 =a2+2ab+b2
⇒ x2 – x – 2 = x2 +2x – 3
⇒ 3x – 1 = 0
Since the above equation is not in the form ofax2 + bx + c = 0.
Therefore, the given equation is not aquadratic equation.
(iv) Given, (x – 3)(2x +1) = x(x + 5)
By using the formula for (a+b)2=a2+2ab+b2
⇒ 2x2 – 5x – 3 = x2 +5x
⇒ x2 – 10x – 3 = 0
Since the above equation is in the form of ax2 + bx + c =0.
Therefore, the given equationis quadratic equation.
(v) Given, (2x – 1)(x – 3) =(x + 5)(x – 1)
By using the formula for (a+b)2=a2+2ab+b2
⇒ 2x2 – 7x + 3 = x2 +4x – 5
⇒ x2 – 11x + 8 = 0
Since the above equation is in the form of ax2 + bx + c =0.
Therefore, the given equationis quadratic equation.
(vi) Given, x2 + 3x + 1 =(x – 2)2
By using the formula for (a+b)2=a2+2ab+b2
⇒ x2 + 3x + 1 = x2 +4 – 4x
⇒ 7x – 3 = 0
Since the above equation is not in the form ofax2 + bx + c = 0.
Therefore, the given equation is not aquadratic equation.
(vii) Given, (x + 2)3 =2x(x2 – 1)
By using the formula for (a+b)2 =a2+2ab+b2
⇒ x3 + 8 + x2 +12x = 2x3 – 2x
⇒ x3 + 14x – 6x2 – 8 =0
Since the above equation is not in the form ofax2 + bx + c = 0.
Therefore, the given equation is not aquadratic equation.
(viii) Given, x3 – 4x2 – x +1 = (x – 2)3
By using the formula for (a+b)2 =a2+2ab+b2
⇒ x3 – 4x2 – x +1 = x3 – 8 – 6x2 + 12x
⇒ 2x2 – 13x + 9 = 0
Since the above equation is in the form of ax2 + bx + c =0.
Question - 9 : - Represent the following situations in the form of quadratic equations:
Answer - 9 : -
(i) The area of arectangular plot is 528 m2. The length of the plot (in metres) isone more than twice its breadth. We need to find the length and breadth of theplot.
(ii) The product oftwo consecutive positive integers is 306. We need to find the integers.
(iii) Rohan’s motheris 26 years older than him. The product of their ages (in years) 3 years fromnow will be 360. We would like to find Rohan’s present age.
(iv) A train travelsa distance of 480 km at a uniform speed. If the speed had been 8 km/h less, thenit would have taken
Solutions:
(i) The area of a rectangular plot is 528 m_2. 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.
Solution
(i) Let us consider,
Breadth of the rectangular plot= x m
Thus, the length of the plot = (2x + 1)m.
As we know,
Area of rectangle = length × breadth= 528 m2
Putting the value of length and breadth of theplot in the formula, we get,
(2x + 1) × x = 528
⇒ 2x2 + x =528
⇒ 2x2 + x – 528 = 0
Therefore, the length and breadth of plot,satisfies the quadratic equation, 2x2 + x – 528 = 0,which is the required representation of the problem mathematically.
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
Solution
(ii) Let us consider,
The first integer number = x
Thus, the next consecutive positive integerwill be = x + 1
Product of two consecutive integers= x × (x +1) = 306
⇒ x2 + x = 306
⇒ x2 + x – 306 = 0
Therefore, the two integers x and x+1,satisfies the quadratic equation, x2 + x – 306 = 0,which is the required representation of the problem mathematically.
(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
Solution
(iii) Let us consider,
Age of Rohan’s = x years
Therefore, as per the given question,
Rohan’s mother’s age = x + 26
After 3 years,
Age of Rohan’s = x + 3
Age of Rohan’s mother will be = x +26 + 3 = x + 29
The product of their ages after 3 years willbe equal to 360, such that
(x + 3)(x + 29) = 360
⇒ x2 + 29x + 3x + 87 = 360
⇒ x2 + 32x + 87 – 360 = 0
⇒ x2 + 32x – 273 = 0
Therefore, the age of Rohan and his mother,satisfies the quadratic equation, x2 + 32x – 273 = 0,which is the required representation of the problem mathematically.
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken
Solution
(iv) Let us consider,
The speed of train = x km/h
And
Time taken to travel 480 km = 480/x km/hr
As per second condition, the speed of train =(x – 8) km/h
Also given, the train will take 3 hours tocover the same distance.
Therefore, time taken to travel 480 km =480/(x+3) km/h
As we know,
Speed × Time = Distance
Therefore,
(x – 8)(480/(x + 3) =480
⇒ 480 + 3x – 3840/x – 24 = 480
⇒ 3x – 3840/x = 24
⇒ 3x2 – 8x – 1280 = 0
Therefore, the speed of the train, satisfiesthe quadratic equation, 3x2 – 8x – 1280 = 0,which is the required representation of the problem mathematically.
Question - 10 : - Find the roots of the following quadratic equations by factorisation:
Answer - 10 : -
(i) x2 –3x – 10 = 0
(ii) 2x2 + x – 6 = 0
(iii) √2 x2 + 7x + 5√2 = 0
(iv) 2x2 – x +1/8 = 0
(v) 100x2 – 20x + 1 = 0
Solutions:
(i) Given, x2 – 3x –10 =0
Taking LHS,
=>x2 – 5x +2x – 10
=>x(x – 5) + 2(x –5)
=>(x – 5)(x + 2)
The roots of this equation, x2 –3x – 10 = 0 are the values of x for which (x – 5)(x + 2)= 0
Therefore, x – 5 = 0 or x +2 = 0
=> x = 5 or x =-2
(ii) Given, 2x2 + x –6 = 0
Taking LHS,
=> 2x2 + 4x –3x – 6
=> 2x(x + 2) – 3(x +2)
=> (x + 2)(2x – 3)
The roots of this equation, 2x2 + x –6=0 are the values of x for which (x – 5)(x + 2)= 0
Therefore, x + 2 = 0or 2x – 3 = 0
=> x = -2 or x =3/2
(iii) √2 x2 + 7x +5√2=0
Taking LHS,
=> √2 x2 + 5x +2x + 5√2
=> x (√2x +5) + √2(√2x + 5)= (√2x + 5)(x + √2)
The roots of this equation, √2 x2 +7x + 5√2=0 are the values of x for which (x – 5)(x + 2)= 0
Therefore, √2x + 5 = 0or x + √2 = 0
=> x = -5/√2 or x =-√2
(iv) 2x2 – x +1/8= 0
Taking LHS,
=1/8 (16x2 – 8x +1)
= 1/8 (16x2 – 4x -4x +1)
= 1/8 (4x(4x –1) -1(4x – 1))
= 1/8 (4x – 1)2
The roots of this equation, 2x2 – x +1/8 = 0, are the values of x for which (4x – 1)2= 0
Therefore, (4x – 1) = 0 or (4x –1) = 0
⇒ x = 1/4 or x = 1/4
(v) Given, 100x2 – 20x +1=0
Taking LHS,
= 100x2 – 10x –10x + 1
= 10x(10x – 1) -1(10x – 1)
= (10x – 1)2
The roots of this equation, 100x2 –20x + 1=0, are the values of x for which (10x – 1)2=0
∴ (10x – 1) = 0 or (10x – 1) = 0
⇒x = 1/10 or x = 1/10