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About Math Assignment: Mathematics or maths as it is popularly called is a game of numbers. These numbers work under a set of rules that set the rule for the “number game i.e mathematics�. Maths is by far the easiest subject in the curriculum.
The student needs to know the rules of the
number game and he is a winner if he plays it well. Due to its logical structure maths has began to be counted as the basis for many subjects. Natural sciences(physics , chemistry etc), engineering, medicine , finance and even social science like economics are based very strongly on mathematics. Many students are heard moaning about mathematics being scary or is it just a myth. Well there can be no great myth than the fact that mathematics is tough. It is indeed an apathy to term mathematics a Frankenstein. Success in mathematics is easy and paves the way for success in other subjects. Befriend numbers and they will befriend you back. We Provide math homework, help with math assignment to solve problems about math and scoring good marks in exam. Sample of Math Assignment Illustrations and Solutions: Question-1: If x = 2 is a root of x2+2kx + 4 =0, then k = ‌‌ (a) -1 , (b) -2 , (C) 2 Solution: Since x = 2 is a root of x2 +2kx + 4 = 0 ∴ (2)2 + 2k(2) + 4 = 0 4K + 8 = 0 = K = -2 ∴ (b) holds. Question-2:
If
1 2
5
is a root of the equation x2 + kx - = 0, then the value of k is 4
(a)2 , (b) -2 , (c)
1 4
1
, (d) 2
Solution: Since x =
1 2
is a root ∴
1 4
+
đ?‘˜ 2
5
-4 = 0 =
đ?‘˜ 2
- 1=0=k=2
∴ (a) holds.
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Question-3: Find the roots of the following quadratic equations by factorization method : 6x2 - 2x – 2 =0 4 3� 2 +5x -2 3 = 0 10a� 2 -6x+15ax -9 = 0, a ≠0
(i) (ii) (iii) Solution:
(i) 6x2 - 2 x-2 = 6đ?‘Ľ 2 - 3 2 x + 2 = 3đ?‘Ľ +
2đ?‘Ľ − 2 ∴ 6đ?‘Ľ 2 - 2đ?‘Ľ - 2 = 0 = 3đ?‘Ľ +
2
= 3đ?‘Ľ + 2
=-
2 3
and
2đ?‘Ľ -2 = 3x (2x - 2) + 2 2đ?‘Ľ − 2
2đ?‘Ľ − 2 = 0 = x =2 2
2 3
or x =
2 2đ?‘Ľ − 2 = 0
2 2
are roots of the given quadratic equation. [âˆľ 4 3 (-2 3) = -24 = 8 Ă— -3]
(ii) 4 3đ?‘Ľ 2 + 5x -2 3 đ?‘Ľ 2 + 8x -3x -2 3 =0 = 4x
3đ?‘Ľ + 2 - 3 [ 3x + 2] = 0 =
3đ?‘Ľ + 2
4đ?‘Ľ − 3 =0
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=
3x + 2 = 0 or 4x - 3 = 0 = x =-
2 3
or x =
3 . 4
(iii)10ax2 – 6x + 15ax – 9 = 0 = 2x (5ax-3) + 3 (5ax-3) =0 = (5ax-3) (2x+ 3) = 0 = 5ax-3 or 2x + 3 = 0 = 5 ax = 3 or 2x = -3 = x =
Question-4:
3 5đ?‘Ž
or x =-
3 2
Solve: a2b2x2 +b2x -1 = 0
Solution: a2b2x2+b2x-a2x - đ?‘Ž2 x -1 = 0 = đ?‘? 2 x [đ?‘Ž2 x + 1) = 0 = đ?‘Ž2 đ?‘Ľ + 1
đ?‘? 2 đ?‘Ľ − 1 = 0 = đ?‘Ž2 x + 1 = 0 or đ?‘? 2 x – = 0 =x =
1 đ?‘Ž2
or x =
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1 đ?‘?2
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Solve for x : 12 abx2 - 9đ?‘Ž2 − 8đ?‘? 2 x – 6ab = 0
Question-5: Solution:
12abx2 – (9đ?‘Ž2 − 8đ?‘? 2 ) x – 6ab = 0 = 12 abx2 – 9a2x + 8đ?‘? 2 x-6ab = 0 = 4bx – 3a = 0 or 3ax + 2b = 0 = x = Question-6: Solve :
1 đ?‘Ž+đ?‘?+đ?‘Ľ
=
1
+
đ?‘Ž
1 đ?‘?
3đ?‘Ž
or x =
4đ?‘?
+
1 đ?‘Ľ
2đ?‘? 3đ?‘Ž
.
: a ≠0 b ≠0, x ≠0
Solution: đ?&#x;? đ?’‚+đ?’ƒ+đ?’™
=
=
đ?&#x;?
+ đ?’‚
−(đ?‘Ž+đ?‘? (đ?‘Ľ 2 +đ?‘Žđ?‘Ľ +đ?‘?đ?‘Ľ
đ?&#x;?
+ đ?’ƒ
đ?&#x;?
=
đ?‘Ž+đ?‘?
đ?’™
=
đ?‘Žđ?‘?
1 đ?‘Ž +đ?‘? +đ?‘Ľ
=
-
1 đ?‘Ľ
=
1 đ?‘Ž
−1 đ?‘Ľ 2 +đ?‘Žđ?‘Ľ +đ?‘?đ?‘Ľ
+
1 đ?‘?
=
=
đ?‘Ľâˆ’đ?‘Ž −đ?‘?−đ?‘Ľ đ?‘Ž +đ?‘?+đ?‘Ľ đ?‘Ľ
=
đ?‘?+đ?‘Ž đ?‘Žđ?‘?
1 đ?‘Žđ?‘?
= x2 +ax + bx = -ab = x (x+a) +b(x+a) = 0 = (x + a ) (x + b) = 0 = x = -a or x = -b
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Question-7: = 2đ?‘Ľ − đ?‘Ž2
2
Solve for x : 4đ?‘Ľ 2 - 4đ?‘Ž2 x + đ?‘Ž4 − đ?‘? 4 = 0
- đ?‘?2
2
= 0 [âˆľ đ?‘Ž2 -đ?‘? 2 = đ?‘Ž − đ?‘?
đ?‘Ž+đ?‘? ]
= (2x – đ?‘Ž2 -đ?‘? 2 ) (2x - đ?‘Ž2 + đ?‘? 2 ) = 0 = 2x - đ?‘Ž2 -đ?‘? 2 = o or 2x -đ?‘Ž2 + đ?‘? 2 = 0 = 2x = đ?‘Ž2 + đ?‘? 2 = 0 or 2x = đ?‘Ž2 − đ?‘? 2 = x = Question-8: Solve:
đ?‘‹+1 đ?‘‹âˆ’1
đ?‘‹âˆ’2
+
đ?‘‹+2
đ?‘Ž 2 +đ?‘? 2 2
or = x =
đ?‘Ž 2 −đ?‘? 2 2
= 3 ; x ≠1, -2
Solution: (i) =
=
đ?‘Ľ+1 đ?‘Ľ+2 đ?‘Ľâˆ’1 (đ?‘Ľâˆ’2) đ?‘Ľâˆ’đ?‘Ž (đ?‘Ľ+2)
2đ?‘Ľ 2 +4
=3=
đ?‘Ľ 2 + 3đ?‘Ľ+2)+(đ?‘Ľ 2 −3đ?‘Ľ+2 đ?‘Ľâˆ’1 (đ?‘Ľ+2)
=3
= 3 = 2� 2 + 4 = 3� 2 + 3x – 6 = 0 = � 2 + 3x -10
đ?‘‹ 2 +đ?‘Ľâˆ’2
= 0 = � 2 + 5x – 2x – 10 = x(x + 5 ) = 0 = (x+5) (x-2) = x = -5 or x = 2 Question-9: Solve for x :
đ?‘Ľâˆ’1
+
đ?‘Ľâˆ’2
đ?‘Ľâˆ’2 đ?‘Ľ+2
=
10 3
(� ≠2, x ≠4)
Solution: We have :
=
đ?‘Ľâˆ’1 đ?‘Ľâˆ’2
+
đ?‘Ľâˆ’3 đ?‘Ľâˆ’4
đ?‘Ľ 2 − 5đ?‘Ľ+4 + đ?‘Ľ 2 −5đ?‘Ľ+6 đ?‘Ľ 2 − 6đ?‘Ľ+8
=
10 3
=
=
10 3
=
đ?‘Ľ −1 đ?‘Ľâˆ’4 + đ?‘Ľâˆ’2 đ?‘Ľâˆ’3 đ?‘Ľâˆ’2 đ?‘Ľâˆ’4 2đ?‘Ľ 2 − 10đ?‘Ľ+10 đ?‘Ľ 2 − 6đ?‘Ľ+8
=
=
10 3
10 3
= 10� 2 -60x + 80 = 6� 2 -30x + 30 = 4� 2 -30x + 50 = 0 = 2� 2 -15x + 25 =0 = 2� 2 -10x – 5x + 25 = 0 = 2x (x-5) -5 (x-5) = 0 = 2� 2 - 15x + 25 = 0 = 2� 2 -10 -5x + 25 = 0 = 2x (x-5) -5 (x-5) = 0 = (x-5) (2x – 5) = 0 = Either x-5 = 0 or 2x -5 = 0 = x = 5 or x = Hence, the solution are 5 and
5 2
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5 2
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Question-10: Solve the following quadratic equations by factorization method : 4 5 3 2đ?‘Ľ (i) - 3 = ; x ≠0, - (ii) đ?‘Ľ 2đ?‘Ľ+3 2 đ?‘Ľâˆ’3
+
1 2đ?‘Ľ+3
+
3đ?‘Ľ+9 đ?‘Ľâˆ’3 2đ?‘Ľ+3
= 0 (C.B.S.E. 2006 C)
Solution: (i) We have
4 đ?‘Ľ
-
3=
5 2đ?‘Ľ+3
=
4−3đ?‘Ľ đ?‘Ľ
=
5 2đ?‘Ľ+3
= (4-3x) (2x + 3) = 5x = 12 –x -6� 2 +6x – 12 = 0 = � 2 + x -2 = 0 = � 2 + 2x –x -2 = 0 =x(x +2) – (x +2) = 0 = � + 2
đ?‘Ľâˆ’1 = 0
= x + 2 =0 or x-1 = 0 = x = -2 or x = 1 3
(ii) Clearly, the given equation is valid if x-3 ≠0 and 2x + 3 ≠0 i.e., when x ≠- , 3 2
Now,
2đ?‘Ľ đ?‘Ľâˆ’3
+
1 2đ?‘Ľ+3
+
3đ?‘Ľ+9 đ?‘Ľâˆ’3 2đ?‘Ľ+3
=0
=2x 2đ?‘Ľ + 3 + đ?‘Ľ − 3 + 3x + 9 = 0 [Multiplying throughout by (x-3) (2x + 3)] = 4đ?‘Ľ 2 + 6x + x -3 + 3x + 9 = 0 = 4đ?‘Ľ 2 + 10x + 6 = 0 =
2x2 + 5x + 3 = 0 = 2x2 + 2x + 3x + 3 = 0
= 2x (x+1) + 3 (x+1) = 0 = 2đ?‘Ľ + 3
đ?‘Ľ + 1 = 0 =x + 1 = 0 = x = -1 [âˆľ 2x + 3 ≠0]
Hence, x = -1 is the only solution of the given equation.
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