Quadratic Equations Ex 4.2 Solutions
Question - 1 : - State whether the following quadratic equations have two distinct real roots. Justify your answer.
Answer - 1 : -
- x2 – 3x + 4 = 0
- 2x2 + x – 1 = 0
- 2x2 – 6x + 9/2 = 0
- 3x2 – 4x + 1 = 0
- (x + 4)2 – 8x = 0
- (x – √2)2 – 2(x + 1) = 0
- √2 x2 –(3/√2)x + 1/√2 = 0
- x (1 – x) – 2 = 0
- (x – 1) (x + 2) + 2 = 0
- (x + 1) (x – 2) + x = 0
Solution:
Q i. x2 –3x + 4 = 0
Solution:-
(i)
The equation x2 – 3x + 4 = 0 hasno real roots.
D = b2 – 4ac
= (-3)2 – 4(1)(4)
= 9 – 16 < 0
Hence, the roots are imaginary.
Q ii. 2x2 +x – 1 = 0
Solution:-
(ii)
The equation 2x2 + x – 1 = 0has two real and distinct roots.
D = b2 – 4ac
= 12 – 4(2) (-1)
= 1 + 8 > 0
Hence, the roots are real and distinct.
Q iii. 2x2 –6x + 9/2 = 0
Solution:-
(iii)
The equation 2x2 – 6x +(9/2) = 0 has real and equal roots.
D = b2 – 4ac
= (-6)2 – 4(2) (9/2)
= 36 – 36 = 0
Hence, the roots are real and equal.
Q iv. 3x2 – 4x + 1 = 0
Solution:-
(iv)
The equation 3x2 – 4x + 1 = 0has two real and distinct roots.
D = b2 – 4ac
= (-4)2 – 4(3)(1)
= 16 – 12 > 0
Hence, the roots are real and distinct.
Q v. (x + 4)2 – 8x = 0
Solution:-
(v)
The equation (x + 4)2 – 8x = 0has no real roots.
Simplifying the above equation,
x2 + 8x + 16 – 8x = 0
x2 + 16 = 0
D = b2 – 4ac
= (0) – 4(1) (16)< 0
Hence, the roots are imaginary.
Q vi. (x – √2)2 –2(x + 1) = 0
Solution:-
(vi)
The equation (x – √2)2 – √2(x+1)=0 hastwo distinct and real roots.
Simplifying the above equation,
x2 – 2√2x + 2 – √2x – √2 = 0
x2 – √2(2+1)x + (2 – √2) = 0
x2 – 3√2x + (2 – √2) = 0
D = b2 – 4ac
= (– 3√2)2 – 4(1)(2 – √2)
= 18 – 8 + 4√2 > 0
Hence, the roots are real and distinct.
Q vii. √2 x2 –(3/√2)x+ 1/√2 = 0
Solution:-
(vii)
The equation √2x2 – 3x/√2 + ½= 0 has two real and distinct roots.
D = b2 – 4ac
= (- 3/√2)2 – 4(√2) (½)
= (9/2) – 2√2 > 0
Hence, the roots are real and distinct.
Q viii. x (1 – x) – 2 = 0
Solution:-
(viii)
The equation x (1 – x) – 2 = 0 has no realroots.
Simplifying the above equation,
x2 – x + 2 = 0
D = b2 – 4ac
= (-1)2 –4(1)(2)
= 1 – 8 < 0
Hence, the roots are imaginary.
Q ix. (x – 1) (x + 2) + 2 = 0
Solution:-
(ix)
The equation (x – 1) (x + 2) + 2 = 0 has tworeal and distinct roots.
Simplifying the above equation,
x2 – x + 2x – 2 + 2 = 0
x2 + x = 0
D = b2 – 4ac
= 12 –4(1)(0)
= 1 – 0 > 0
Hence, the roots are real and distinct.
Q x. (x + 1) (x – 2) + x = 0
Solution:-
(x)
The equation (x + 1) (x – 2) + x = 0 has tworeal and distinct roots.
Simplifying the above equation,
x2 + x – 2x – 2 + x = 0
x2 – 2 = 0
D = b2 – 4ac
= (0)2 – 4(1) (-2)
= 0 + 8 > 0
Hence, the roots are real and distinct.
Answer - 2 : -
- Every quadratic equation has exactly one root.
- Every quadratic equation has at least one real root.
- Every quadratic equation has at least two roots.
- Every quadratic equations has at most two roots.
- If the coefficient of x2 and the constant term of a quadratic equation have opposite signs, then the quadratic equation has real roots.
- If the coefficient of x2 and the constant term have the same sign and if the coefficient of x term is zero, then the quadratic equation has no real roots.
Solution:
(i)False. For example, a quadratic equation x2 – 9 = 0has two distinct roots – 3 and 3.
(ii) False. For example, equation x2 +4 = 0 has no real root.
(iii) False. For example, a quadraticequation x2 – 4x + 4 = 0 has only oneroot which is 2.
(iv) True, because every quadratic polynomialhas almost two roots.
(v) True, because in this case discriminant isalways positive.
For example, in ax2+ bx + c =0, as a and c have opposite sign, ac <0
⟹ Discriminant = b2 – 4ac >0.
(vi) True, because in this case discriminantis always negative.
For example, in ax2+ bx + c =0, as b = 0, and a and c havesame sign then ac > 0
⟹ Discriminant = b2 – 4ac =– 4 a c < 0
Question - 3 : - Find the roots of the following quadratic equations by factorisation
Answer - 3 : -
(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
Question - 4 : - Solve the problems given in Example 1.
Represent the following situations mathematically:
Answer - 4 : -
(i) John and Jivantitogether have 45 marbles. Both of them lost 5 marbles each, and the product ofthe number of marbles they now have is 124. We would like to find out how manymarbles they had to start with.
(ii) A cottageindustry produces a certain number of toys in a day. The cost of production ofeach toy (in rupees) was found to be 55 minus the number of toys produced in aday. On a particular day, the total cost of production was ` 750. We would liketo find out the number of toys produced on that day.
Solutions:
(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.
Soiution
(i) Let us say, the number of marbles Johnhave = x.
Therefore, number of marble Jivanti have = 45– x
After losing 5 marbles each,
Number of marbles John have = x –5
Number of marble Jivanti have = 45 – x –5 = 40 – x
Given that the product of their marbles is124.
∴ (x – 5)(40 – x) = 124
⇒ x2 – 45x + 324 = 0
⇒ x2 – 36x – 9x +324 = 0
⇒ x(x – 36) -9(x – 36) = 0
⇒ (x – 36)(x – 9) = 0
Thus, we can say,
x – 36 = 0or x – 9 = 0
⇒ x = 36 or x = 9
Therefore,
If, John’s marbles = 36,
Then, Jivanti’s marbles = 45 – 36 = 9
And if John’s marbles = 9,
Then, Jivanti’s marbles = 45 – 9 = 36
(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 ` 750. We would like to find out the number of toys produced on that day.
Solution
(ii) Let us say, number of toys produced in aday be x.
Therefore, cost of production of each toy =Rs(55 – x)
Given, total cost of production of the toys =Rs 750
∴ x(55 – x) = 750
⇒ x2 – 55x + 750 = 0
⇒ x2 – 25x – 30x +750 = 0
⇒ x(x – 25)-30(x – 25) = 0
⇒ (x – 25)(x – 30) = 0
Thus, either x -25 = 0or x – 30 = 0
⇒ x = 25 or x = 30
Hence, the number of toys produced in a day,will be either 25 or 30.
Question - 5 : - Find two numberswhose sum is 27 and product is 182.
Answer - 5 : -
Let us say, first number be x andthe second number is 27 – x.
Therefore, the product of two numbers
x(27 – x) = 182
⇒ x2 – 27x – 182 = 0
⇒ x2 – 13x – 14x + 182 = 0
⇒ x(x – 13) -14(x – 13) = 0
⇒ (x – 13)(x -14) = 0
Thus, either, x = -13 = 0or x – 14 = 0
⇒ x = 13 or x = 14
Therefore, if first number = 13, then secondnumber = 27 – 13 = 14
And if first number = 14, then second number =27 – 14 = 13
Hence, the numbers are 13 and 14.
Question - 6 : - Find two consecutivepositive integers, sum of whose squares is 365.
Answer - 6 : -
Let us say, the two consecutive positiveintegers be x and x + 1.
Therefore, as per the given questions,
x2 + (x + 1)2 = 365
⇒ x2 + x2 +1 + 2x = 365
⇒ 2x2 + 2x – 364 = 0
⇒ x2 + x – 182 = 0
⇒ x2 + 14x – 13x –182 = 0
⇒ x(x + 14) -13(x + 14)= 0
⇒ (x + 14)(x – 13) = 0
Thus, either, x + 14 = 0or x – 13 = 0,
⇒ x = – 14 or x = 13
since, the integers arepositive, so x can be 13, only.
∴ x + 1 = 13 + 1 = 14
Therefore, two consecutive positive integerswill be 13 and 14.
Question - 7 : - The altitude of aright triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find theother two sides.
Answer - 7 : -
Let us say, the base of the right trianglebe x cm.
Given, the altitude of right triangle =(x – 7) cm
From Pythagoras theorem, we know,
Base2 + Altitude2 =Hypotenuse2
∴ x2 + (x – 7)2 = 132
⇒ x2 + x2 + 49 –14x = 169
⇒ 2x2 – 14x – 120 = 0
⇒ x2 – 7x – 60 = 0
⇒ x2 – 12x + 5x – 60 = 0
⇒ x(x – 12) + 5(x – 12) = 0
⇒ (x – 12)(x + 5) = 0
Thus, either x – 12 = 0or x + 5 = 0,
⇒ x = 12 or x = – 5
Since sides cannot benegative, x can only be 12.
Therefore, the base of thegiven triangle is 12 cm and the altitude of this triangle will be (12 – 7) cm =5 cm.
Question - 8 : - A cottage industryproduces a certain number of pottery articles in a day. It was observed on aparticular day that the cost of production of each article (in rupees) was 3more than twice the number of articles produced on that day. If the total costof production on that day was Rs.90, find the number of articles produced andthe cost of each article.
Answer - 8 : -
Let us say, the number of articles producedbe x.
Therefore, cost of production of each article= Rs (2x + 3)
Given, total cost of production is Rs.90
∴ x(2x + 3) = 90
⇒ 2x2 + 3x – 90 = 0
⇒ 2x2 + 15x -12x –90 = 0
⇒ x(2x + 15) -6(2x + 15) = 0
⇒ (2x + 15)(x – 6) = 0
Thus, either 2x + 15 = 0 or x –6 = 0
⇒ x = -15/2 or x = 6
As the number of articles produced can only bea positive integer, therefore, x can only be 6.
Hence, number of articles produced = 6
Cost of each article = 2 × 6 + 3 = Rs 15.