RD Chapter 8 Quadratic Equations Ex 8.7 Solutions
Question - 11 : - The sum of a number and its square is 63/4 , find the numbers.
Answer - 11 : -
Question - 12 : - There are three consecutive integers such that the square of the first increased by the product of the other two gives 154. What are the integers ?
Answer - 12 : -
Let first integer = x
Then second integer = x + 1
and third integer = x + 2
According to the condition,
x² + (x + 1) (x + 2) = 154
=> x² + x² + 3x + 2 = 154
=> 2x² + 3x + 2 – 154 = 0
=> 2x² – 16x + 19x – 152 = 0
=> 2x(x – 8) + 19 (x – 8) = 0
=> (x – 8) (2x + 19) = 0
Either x – 8 = 0, then x = 8
or 2x + 19 = 0, then 2x = -19 => x = (−19/2) But it is not an integer
First number = 8
Second number = 8 + 1=9
and third number = 8 + 2 = 10
Question - 13 : - The product of two successive integral multiple of 5 is 300. Determine the multiplies.
Answer - 13 : -
Let first multiplie of 5 = 5x
Then second multiple = 5x + 5
According to the condition,
5x (5x + 5) = 300
=> 25 x² + 25x – 300 = 0
=> x² + x – 12 = 0 (Dividing by 25)
=> x² + 4x – 3x – 12 = 0
=> x (x + 4) – 3 (x + 4) = 0
=> (x – 4) (x – 3) = 0
Either x + 4 = 0, then x = -4
or x – 3 = 0, then x = 3
(i) When x = -4, then
Required multiples of 5 will be
5 (-4) = -20, -20 + 5 = -15
or when x = 3, then
Required multiples will be
5 x 3 = 15, 15 + 5 = 20
Required number are -20, -15 or 15, 20
Question - 14 : - The sum of the squares of two numbers is 233 and one of the numbers is 3 less than twice the other number. Find the nqmbers.
Answer - 14 : -
Let first number = x
Then second number = 2x – 3
According to the condition,
x² + (2x – 3)2 = 233
=> x² + 4x² – 12x + 9 = 233
=> 5x² – 12x + 9 – 233 = 0
=> 5x² – 12x – 224 = 0
=> 5x² – 40x + 28x – 224 = 0
=> 5x (x – 8) + 28 (x – 8) = 0
=> (x – 8) (5x + 28) = 0
Either x – 8 = 0, then x = 8
or 5x + 28 = 0, then 5x = -28 => x = −28/5 But it is not possible
x = 8
First number = 8
Second number = 2x – 3 = 2 x 8 – 3 = 16 – 3 = 13
Number are 8, 13
Question - 15 : - Find two consecutive even integers whose squares have the sum 340.
Answer - 15 : -
Let first even integer = x
The second even integer = x + 2
According to the condition,
x² + (x + 2)2 = 340
x² + x² + 4x + 4 = 340
=> 2x² + 4x + 4 – 340 = 0
=> 2x² + 4x – 336 = 0
=> x² + 2x – 168 = 0
=> x² + 14x – 12x – 168 = 0
=> x (x + 14) – 12 (x + 14) = 0
=> (x + 14) (x – 12) = 0
Either x + 14 = 0, then x = -14
or x – 12 = 0, the x = 12
(i) If x = -14, then
First number = -14
and second number = -14 + 2 = -12
(ii) If x = 12, then
First number =12
and second number =12 + 2 = 14
Hence even numbers are 12, 14 or -14, -12
Question - 16 : - The difference of two numbers is 4. If the difference of their reciprocals is 4/21, find the numbers. (C.B.S.E. 2008)
Answer - 16 : -
Let first number = x
Then second number = x – 4
According to the condition,
Either x – 7 = 0, then x = 7
or x + 3 = 0, then x = -3
(i) If x = 7, then
First number = 7
and second number = 7 – 4 = 3
(ii) If x = -3, then
First number = -3
and second number = -3 – 4 = -7
Number are 7, 3 or -3, -7
Question - 17 : - Find two natural numbers which differ by 3 and whose squared have the sum 117.
Answer - 17 : -
Let first number = x
Then second number = x – 3
According to the condition,
x² + (x – 3)2 = 117
=> x² + x² – 6x + 9 = 117
=> 2x² – 6x + 9 – 117 = 0
=> 2x² – 6x – 108 = 0
=> x² – 3x – 54 = 0 (Dividing by 2)
=> x² – 9x + 6x – 54 = 0
=> x (x – 9) + 6 (x – 9) = 0
=> (x- 9) (x + 6) = 0
Either x – 9 = 0, then x = 9
or x + 6 = 0, then x = -6 which is not a natural number
First natural number = 9
and second number = 9 – 3 = 6
Question - 18 : - The sum of squares of three consecutive natural numbers is 149. Find the numbers.
Answer - 18 : -
Let first number = x
Then second number = x + 1
and third number = x + 2
According to the condition,
x² + (x + 1)2 + (x + 2)2 = 149
=> x² + x² + 2x + 1 + x2 + 4x + 4 = 149
=> 3x² + 6x + 5 – 149 = 0
=> 3x² + 6x – 144 = 0
=> x² + 2x – 48 = 0 (Dividing by 3)
=> x² + 8x – 6x – 48 = 0
=> x (x + 8) – 6 (x + 8) = 0
=> (x + 8) (x – 6) = 0 .
Either x + 8 = 0, then x = -8, But it is not a natural number
or x – 6 = 0, then x = 6
Numbers are 6, 6 + 1 = 7, 6 + 2 = 8 or 6, 7, 8
Question - 19 : - The sum of two numbers is 16. The sum of their reciprocals is 1/3. Find the numbers. (C.B.S.E. 2005)
Answer - 19 : -
Sum of two numbers = 16
Let first number = x
Then second number = 16 – x
According to the condition,
Either x – 12 = 0, then x = 12
or x – 4 = 0, then x = 4
(i) If x = 12, then
First number = 12
and second number = 16 – 12 = 4
(ii) If x = 4, then First number = 4
and second number = 16 – 4 = 12
Hence numbers are 4, 12
Question - 20 : - Determine two consecutive multiples of 3 whose product is 270.
Answer - 20 : -
Let first multiple of 3 = 3x
Then second multiple of 3 = 3x + 3
According to the condition,
3x (3x + 3) = 270
=> 9x² + 9x – 270 = 0
=> x² + x – 30 = 0 (Dividing by 9)
=> x² + 6x – 5x – 30 = 0
=> x (x + 6) – 5 (x + 6) = 0
=> (x + 6) (x – 5) = 0
Either x + 6 = 0, then x = -6
or x – 5 = 0, then x = 5
(i) When x = -6, then
First number = 3x = 3 x (-6) = -18 and second number = -18 + 3 = -15
(ii) If x = 5, then
First number = 3x = 3 x 5 = 15 and second number =15 + 3 = 18
Hence numbers are 15, 18 or -18, -15