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Sample of Math Homework Help Illustrations and Solutions:
Example: If the root of the equation 2𝑥2
- 10x + 𝜌 = 0 = is 2, then the value of 𝜌 is :
(a) -3 (b) -6, (c) 9 (d) 12
Solution. Since is a root of the equation
∴ 2(2)2
- 10(2) + 𝜌 =0 = 𝜌 = 20 – 8 = 12 ∴ (d) holds.
Example: Which of the following is a root of the equation 2𝑥2
- 5x – 3 = 0 ?
(a) x = 3 (b) x = 4 (c) x = 1 (d) x = -4
Solution. (a) holds. (∵ 2(3)2
-5(3) – 3 = 0 = 18 – 15 – 3 = 0)
Example: If no root of 𝑥2
-kx + 1 = 0 is real, then
(a) -3 < k < 3 (b) –2 <k <2 (c) k > 2 (d) k < -2
(b) Solution. Since the roots of 𝑥2
– kx + 4 = 0 are non-real.
∴ Disc. (−𝑘)2
- 4 < 0 = 𝑘2
- 4 < 0 = 𝑘2
< 4 = |k| <2 ∴ (b) holds
Example: If 𝑥2
+ 4x + k = 0 has real roots, then
(a) -3 < k < 3 (b) -2 <k <2 (c) k > 2 (d) k < -2
(C.B.S.E. 2012)
Solution. Since the roots of 𝑥2
- kx + 4 = 0 are non-real.
∴ Disc. (−𝑘)2
-4 < 0 = 𝑘2
< 4 = |k| < 2 = -2 < k < 2 ∴ (b) holds.
Example: If 𝑥2
+ 4x + k = 0 has real roots, then

Homework1
(a) k ≥ 4 (b) k ≤ 4 (c) k ≤ 0 (d) k ≥ 0
Solution. Since 𝑥2
+ 4x + k = has real roots.
∴ Disc. (4)2
– 4k ≥ 0 = 4k ≥ 16 = k ≤ 4 ∴ (b) holds.
Example: Value of k for which the quadratic equation 2𝑥2
- kx + k = 0 has equal roots is
(a) 0 only (b)4 (c) 8 only (d) 0.8.
(N.C.E.R.T. Exemplar Problem)
Solution. For equal roots, Disc. = (𝑘)2
-4(2) (k) = 0 = k (k – 8) = 0 = k = 0.8
∴ (d) holds.
Example: The value of k for which 3𝑥2
+ 2x + k = 0 has real roots is :
(a) k >
1
3
(b) k ≤
1
3
(c) k ≤
1
3
(d) k <
1
3
Solution. For real roots, Disc. = (2)2
– 4(3) (k) ≥ 0
= 4 – 12k ≥ 0 = 4 ≥ 12 k = 12 k ≤ 4 = k ≤
4
12
=
1
13
= ∴ (b) holds.
Example: If the quadratic equation 𝑚𝑥2
+ 2x + m = 0 has two equal roots, then the
values of m are
(a) ± 1 (b) 0,-2 (c) 0,1 (d) -1,0

Homework1
Solution. Disc. = 𝑏2
- 4 (6) (2) = 1 (given) = 𝑏2
= 48 + 1 = 49 =b = ± 7 ∴ (c) holds.
Solved examples – 6
Example: Find the sum and product of roots of (-2)𝑥2
+ 5x + 4 = 0.
Solution. Sum of roots =
−𝑏
𝑎
=
−5
−2
=
5
2
; product of roots =
𝑐
𝑎
=
4
−2
=-2
Example: If 𝛼 and 𝛽 are roots of 9𝑥2
- 24x + 8 = 0, then find 𝛼 + 𝛽 and 𝛼𝛽.
Solution. 𝛼 + 𝛽 =
−𝑏
𝑎
=
−(−24)
9
=
8
3
; 𝛼𝛽 =
𝑐
𝑎
=
8
9
Example: Form a quadratic equation roots -
1
3
and
5
2
.
Solution. S = Sum of roots = −
1
3
+
5
2
=
−2+15
6
; P = Product of roots = −
1
3
5
2
= -
5
6
∴ equation is 𝑥2
– Sx + p = 0 = 𝑥2
-
13
6
x -
5
6
= 0 = 6𝑥2
-13x – 5 = 0
Example: If one root of quadratic equation with rational co-efficients is 2 + 3 , then
give other root.

Homework1
Solution. Note that in a quadratic equation with rational co-efficients, surd roots occur
in conjugate pairs. = 2 - 3 is the other root.
1. The sum of the squares of two positive integers is 208. If the square of the larger
number is 18 times the smaller, find the numbers.
Solution. Let the smaller number = x ; Square of larger number = 18x
Given, 𝑥2
+ 18x = 208 = 𝑥2
+ 18x – 208 = 0
= (x – 8) (x + 26) = 0 =x = 8, -26 = x = 8
∴ Square of larger number = 18x = (18) (8) = 144
= Larger number = 144 = 12 and smaller number = x = 8
2. If -5 is root of quadratic equation 2𝑥2
+ 2px – 15 = 0 and the quadratic equation
𝝆 (𝑥2
+ x) k = 0 has equal roots, find the value of k.
Solution. 5 is root of 2𝑥2
+ 2px – 15 = 0 = 2 (−5)2
+ 2𝑃 (-5) – 15 = 0 = 𝑃 =
7
2
∴ 𝑃 (𝑥2
+ x) + k = 0 =
7
2
(𝑥2
+ x) + k = 0 = 7𝑥2
+ 7x + 2k = 0
It has equal roots = D = 0 = (7)2
-4(7) (2k) = 0 = k =
7
8
3. If, p, q, r and s are real numbers such that pr = 2(q + s), then show that at least
one of the equations 𝑥2
+ px + q = 0 has real roots.
4. If the roots of the equation 𝑥2
+ 2cx + ab = 0 are real and unequal, prove that
the equation 𝑥2
-2 (a + b) x + 𝑎2
+ 𝑏2
+ 2𝑐2
= 0 has no real roots.
5. A person on tour has $360 for his expenses. If the extends his tour for 4 days, he
has to cut down his daily expenses by $3. Find the original duration of the tour.

Homework1
Solution. Let the original duration of the tour be x days.
∴ total expenditure on tour = $360 ∴ expenditure per day = $
360
𝑥
Duration of the extended tour = (x + 4) days
∴ expenditure per day according to new schedule = $
360
𝑥+4
Since the daily expenses are cut down by $3.
∴
360
𝑥
-
360
𝑥+4
= 3 =
360 𝑥+4 − 360 𝑥
𝑥(𝑥+4)
= 3
=
360 𝑥+1440−360 𝑥
𝑥 (𝑥+4)
= 3 =
1440
𝑥2+ 4𝑥
= 3 = 𝑥2
+ 4x = 480 = 𝑥2
+ 4x – 480 = 0
= 𝑥2
+ 24x – 20x – 480 = 0 = x (x + 24) -20 (x + 24) = 0
= (x – 20) (x + 24) = 0 = x – 20 = 0 or, x + 24 = 0 =x = 20 or, x = -24
But, the number of days cannot be negative. So, x = 20
Hence, the original duration of the tour was of 20 days.
6. (a) A shopkeeper buys a number of books for $80. If he had bought 4 more
books for the same amount, each book would have cost $12 less. How many
books did he buy?

Homework1
(b) A shopkeeper buys a number of books for $1200. If he had bought 10 more books
for the same amount, each book would have cost him $20 less. How many books did he
buy ?
Solution. (a) Let number of books bought be x. Then, Cost of x books = $80
= Cost of one book = $
80
𝑥
If the number of books bought is x + 4, then Cost of one book = $
80
𝑥+4

It is given that the cost of one book is reduced by one rupee
∴
80
𝑥
-
80
𝑥+4
= 1 = 80
1
𝑥
−
1
𝑥+4
= 1 = 80
𝑥+4−𝑥
𝑥 𝑥+4
= 1 =
320
𝑥2+ 4𝑥
= 1
= 𝑥2
+ 4x = 320 = 0 = 𝑥2
+ 20x – 16x – 320 = 0 = x (x + 20) – 16(x + 20) = 0
= (x + 20) (x – 16) = 0 =x = -20 or, x = 16 = x = 16 [∵ x cannot be negative]
Hence, the number of books is 16.
(b) Similar to Part (a) Ans. 20.
7. If the price of a book is reduced by $5, a person can buy 5 more books for $300. Find
the original list price of the book.
Solution. Let the original list price of the book be $ x ∴ number of books bought for
$300 =
300
𝑥
Reduced list price of the book = $(x -5) ∴ number of books bought for $300 =
300
𝑥−5
By the given condition,
300
𝑥−5
-
300
𝑥
= 5 =
300 𝑥−300 𝑥+1500
𝑥(𝑥−5)
= 5 =
1500
𝑥2−5𝑥
= 5
= 𝑥2
- 5x = 300 = 𝑥2
- 5x – 300 = 0 = 𝑥2
- 20x + 15x – 300 = 0 = (x – 20) (x + 15) =
0
= x – 20 = 0 or, x + 15 = 0 = x = 20, x = -15 =x = 20 [∵ x = -15 is not possible]
Hence, the list price of the books = $20
8. A factory kept increasing is doubled in the last two year. Find the percentage if it is
known that the output is doubled in the last two years.
Solution. Let p be the initial production (2 years ago), and let the increase in product
every year be x%. Then,
Product at the end of first year = P +
𝑃𝑥
100
= p 1 +
𝑥
100
Product at the end of the second year
= P 1 +
𝑥
100
+
𝑥
100
1 +
𝑥
100
= P 1 +
𝑥
100
1 +
𝑥
100
= P 1 +
𝑥
100
2

Homework1
Since product is doubled in last two years ∴ P 1 +
𝑥
100
2
= 2P = 1 +
𝑥
100
2
= 2
= (100 + 𝑥)2
= 2 × 1002
= 𝑥2
+ 200x – 10000 = 0
= x =
−200 ± (200)2+ 40000
2
= - 100 ± 100 2 = 100 (-1 + 2)
= x = 100 (-1 + 2) [∵ x cannot be negative]
9. The difference of the ages of Sohrab and his father is 30 years. If the difference of the
squares of their ages is 1560, find their ages.
Solution. Let y, x be the ages of Soharb and his father respectively
∴ x – y = 30 …(1)
and 𝑥2
- 𝑦2
= 1560 …(2)
Divide (2) by (1), we get
𝑥2− 𝑦2
𝑥−𝑦
=
1560
30
= x + y = 52 …(3)
(1) + (2) gives 2x = 82 =x = 41. (3) – (1) gives 2y = 22 = y =
11
Hence required ages are 41 years, 11 years.

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