Chapter-3, Page-51
Exercise 3.1
Problem-5: Express x in terms of a ,b ,c ,d and e
a) a(x – b) + c(x + d) = e
b) b(a + x) + d(c – x) = ex
Solution :
a) a(x – b) + c(x + d) = e
=> ax – ab + cx +cd = e
=> x(a +c) = e + ab –
cd
=> x = (e + ab – cd)/(a
+ c)
b) b(a + x) + d(c – x) = ex
=> ba + bx + dc – dx =
ex
=> x(b – d) = ex – ba
– dc
=> x = (ex – ba –
dc)/(b – d)
Problem-10: Given
that ab – c = d + ae
a) Find a in terms of b , c , d and e
b) Find b in terms of a , c , d and e .
Solution:-
a) ab – c = d + ae
=> ab – ae = d + c
=> a(b – e) = d +
c
=> a = (d +c)/(b – e)
b) ab – c = d + ae
=> ab = ae + c +d
=> b = (ae +c +d)/a
Exercise 3.3, Page 59
Problem: Solve the following
equations:
Solution 5: (3-x) (3x-1)=-11
=>
9x-3-3x2+x+11 = 0
=>
-3x2+10x+8 = 0
=>
3x2-10x-8 = 0
=>
3x2-12x+2x-8 = 0
=>
3x(x-4)+2(x-4) = 0
=>
(x-4)(3x+2) = 0
=>
x-4 = 0 => 3x+2
= 0
=>
x = 4 => 3x
= -2
=>
x = -2/3
Solution 13: (x-3) (x-4)=2
=>
x2-4x-23x+12=0
=>
x2-7x+10=0
=>
x2-5x-2x+10=0
=>
x(x-5)-2(x-5)=0
=>
(x-5)(x-2)=0
=>
x-5=0 =>
x-2=0
=>
x=5 =>
x=2
Solution 15: (x+7) (x-6) = -30
=>
x2-6x+7x-42+30 = 0
=>
x2+x-12 = 0
=>
x2+4x-23x-12 = 0
=>
x(x+4)-3(x+4) = 0
=>
(x+4) (x-3) = 0
=>
x+4 = 0 => x-3 =
0
=>
x = -54 => x =
3
Solution
18: (x-3)2 = 36
=>
x-3 = +6 or -6
=> x = 6+3 => x = -6+3
=> x = 9 => x = -3
Chapter 3, Exercise 3.4, Quadratic
Factorization, Page 63.
Problem 3: Factorize the following:
Solution:
a)x2-x-2
=
x2-2x+x-2
=
x(x-2)+1(x-2)
=
(x-2)(x+1) Ans.
b)
2+x-3x2
= 2+3x-2x-3x2
= 1(2+3x)-x(2+3x)
= (2+3x)(1-x) Ans.
s)
6x2-3-17x
=
6x2-17x-3
=
6x2-18x+x-3
=
6x(x-3)+1(x-3)
=
(x-3)(x+1) Ans.
t)
12x2-18x-12
= 6(2x2-3x-2)
= 6(2x2-4x+x-2)
= 6{2x(x-2)+1(x-2)}
= 6(x-2)(2x+1) Ans
u)
8x-10+2x2
=
2(x2+4x-5)
=
2(x2+5x-x-5)
=
2{x(x+5)-1(x+5)}
=
2(x+5)(x-1) Ans.
v) 12x2-12-32x
= 4(3x2-8x-3)
= 4(3x2-9x+x-3)
= 4{3x(x-3)+1(x-3)}
= 4(x-3)(3x+1) Ans.
Exercise 3.5, Page 65
Problem 2: Solve the following equation:
Solution
(g): (2x+3)(3x-2)+2x+3 = 0
=>
6x2-4x+9x-6+2x+23 = 0
=>
6x2+7x-23 = 0
=>
6x2+9x-2x-3 = 0
=>
3x(2x+3)-1(2x+3) = 0
=>
(2x+3) (x-1) = 0
=>
2x+3 = 0 => x-1 = 0
=>
x = -3/2 => x = 1
Solution
(h): 1+(1-x)(2x+1) = x2
=>
1+2x+1-2x2-x-x2 = 0
=>
-3x2+x+2 = 0
=>
3x2-x-2 = 0
=>
3x2-3x+2x-2 = 0
=>
3x(x-1)+2(x-1) = 0
=>
(x-1)(3x+2) = 0
=>
x-1 = 0 => 3x+2
= 0
=>
x =1 => x
= -2/3
Solution
(i) 3(x-1)2+5x = 5
=>
3(x2-2x+1)+5x-5 = 0
=>
3x2-6x+3+5x-5 = 0
=>
3x2-x-2 = 0
=>
3x2-3x+2x-2 = 0
=>
3x(x-1)+2(x-1) = 0
=>
(x-1)(3x+2) = 0
=>
x-1 = 0 => 3x+2
= 0
=>
x = 1 => x
= -2/3
Solution
(j) 4(x-1)2 – 12(x - 1)
+ 9 = 0
=>
4(x2-2x+1)-(12x+12+9 = 0
=>
4x2-8x+4-12x+21 = 0
=>
4x2-20x=25 = 0
=>
4x2-10x-10x+25 = 0
=>
2x(2x-5)-5(2x-5) = 0
=>
(2x-5)(2x-5) = 0
=>
2x-5 = 0
=>
x = 5/2 Ans.
Chapter-3: 3.6 Word Problems Leading To Quadratic Equations. Page-66, Exercise 3.6
Problem-1: The sum of a number and its square is 156 . Find the number.
Solution :- Let the number be x
According to the question we have
x + x2 = 156
=> x2 + x – 156 = 0
=> x2 + 13x – 12x – 156 = 0
=> x(x + 13) – 12(x + 13) = 0
=> (x + 13) (x – 12) = 0
=> Either x – 12 = 0 or x + 13 = 0 (inadmissible)
=> x = 12 Ans
Problem-7: The area of a triangle is 24 cm2 .If its height is 2 cm longer than its base . Find the base of the triangle .
Solution :- Let the base be x cm . therefore height be x + 2 cm .
According to the question we have
½ x (x +2) = 24
=> x (x + 2) = 48
=> x2 + 2x – 48 = 0
=> x2 + 8x – 6x – 48 = 0
=> x(x + 8) – 6(x + 8) = 0
=> (x + 8) (x – 6) = 0
=> Either x – 6 = 0 or x + 8 = 0 (inadmissible)
=> X = 6 Therefore the base is 6 cm Ans
Problem-8: The sum of the square of two consecutive odd numbers is 290 . Find the two numbers .
Solution :- Let the two odd numbers be 2x + 1 and 2x + 3
According to the question we have
(2x + 1)2 + (2x + 3)2 = 290
=> 4x2 + 4x +1+ 4x2 + 12X +9 – 290 = 0
=> 8x2 + 16x – 280 = 0
=> x2 + 2x – 35 =0
=> x2 + 7x – 5x – 35 = 0
=> x(x + 7) – 5(x + 7) = 0
=> (x + 7) (x – 5) = 0
=> Either x – 5 = 0 or x + 7 = 0 (inadmissible)
=> x = 5 . therefore the odd numbers are 2*5 +1= 11 and 2*5 + 3 = 13 Ans.
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