# Algebra Handwriting Notes In Hindi

### Algebra Handwriting Notes In Hindi PDF Download

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Today, we are sharing ALGEBRA HANDWRITING NOTES IN HINDI. It can prove very useful for upcoming competitive exams like SSC CGL, BANK, RAILWAYS, RRB NTPC, LIC AAO, and many more. So English Grammer Book PDF is very important for any competitive exam. This free PDF will be very helpful for your exam. This PDF is being provided to you for free which you have given below DOWNLOAD button You can do DOWNLOAD by clicking on it, you can also go to the related notes and DOWNLOAD some new PDF related to this PDF. You can learn about all the new updates on by clicking on the Allow button on the screen.

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### Handwritten Algebra Notes in Hindi Main Contents

Algebra

• गुणनखंड (Factor)
• घातांक व करणी (Indices & Surds)
• परिमेय व अपरिमेय संख्याएँ (Rational & Irrational Numbers)
• योगांतरानुपात
• समीकरण (Equations)
• रेखीय समिलारण
• द्विघात समीकरण
• श्री धाराचार्य सूत्र
• मूलों की प्रकृति

### Know About PDF: Handwritten Algebra PDF Notes in Hindi

• Book Name: “Algebra (बीजगणित) PDF Notes
• PDF Size: 6 MB
• No Of Pages: 23 Pages
• Quality: High
• Format: PDF
• language: Hindi/English
• Credit:…..

Math Formula :-

1. (AX+BX) = X(A+B)
2. (X+Y)2  =X2+Y2+2XY
3. (X-Y)2  =X2+Y2-2XY
4. X2-Y= (X+Y)(X-Y)
5. X3+Y3 = (X+Y)(X2-XY+Y2)
6. X3-Y3 = (X-Y)(X2+XY+Y2)
7. (X+Y)3 = X3+Y3+3XY(X+Y)
8. (X-Y)3 = X3-Y3-3XY(X-Y)
9. (X+Y+Z)2 = X2+Y2+Z2+2(XY+YZ+ZX)
10. (X+Y-Z)2 = X2+Y2+Z2+2(XY-YZ-ZX)
11. (X-Y+Z)2 = X2+Y2+Z2+2(-XY-YZ+ZX)
12. (X+Y+Z)= X3+Y3+Z3+3(X+Y)(Y+Z)(Z+X)
13. X3+Y3+Z3-3XYZ = (X+Y+Z)(X2+Y2+Z2-XY-YZ-ZX)

#### Algebra Questions with solutions in Hindi for SSC Exams

Q.1: यदि a + b = p, ab = q है , तो (a4 + b4 ) का मान ज्ञात कीजिए I
(A) p4 – 4p2q + q2
(B) p4 – 4p2q2 + 2q2
(C) p4 – 2p2q2 + q2
(D) p4 – 4p2q + 2q2
(SSC CHSL Aug 2021)

###### 2q + 2q2 a + b = p , ab = q (a + b)2 = p2 a2 + 2ab + b2 = p2 a2 + b2 = p2 – 2q (a2 + b2)2 = (p2 – 2q)2 a4 + b4 + 2a2b2 = p4 – 4p2q + 4q2 [ab = q, a2 b2 = q2] a4 + b4 = p4 – 4p2q + 2q2

Q.2: यदि $(x +\frac 1x)^3 = 27$ है , तो $(x^2 +\frac {1}{x^2})$ का मान क्या होगा ? दिया गया है कि जहां x वास्तविक है I
(A) 9
(B) 25
(C) 7
(D) 11
(SSC CHSL Aug 2021)

Ans : (C) 7 $(x +\frac 1x)^3 = 27$ $x +\frac 1x = (27)^{1/3} = 3$ $x^2 +\frac {1}{x^2} + 2 = 9$ $x^2 +\frac {1}{x^2} = 9 - 2 = 7$

Q.3: यदि $x -\frac 2x = 4$ है , तो $x^2 + \frac {4}{x^2}$ का मान ज्ञात करें I
(A) 18
(B) 8
(C) 12
(D) 20
(SSC CHSL Aug 2021)

Ans : (D) 20 $({x -\frac 2x)^2} = 4$ $x^2 +\frac {4}{x^2} - 2\times2 = 16$ $x^2 +\frac {4}{x^2} = 16 + 4 = 20$

Q.4: यदि $\sqrt{x} + \frac {1} {\sqrt x} = \sqrt 6$ है , तो $x^6 + \frac {1}{x^6}$ का मान ज्ञात करें I
(A) 2712
(B) 2502
(C) 2270
(D) 2702
(SSC CHSL Aug 2021)

Ans : (D) 2702 $\sqrt{x} + \frac {1} {\sqrt x} = \sqrt6$ $(\sqrt{x} + \frac {1} {\sqrt x})^2 = (\sqrt6)^2$ $x +\frac 1x +2 = 6$ $x +\frac 1x = 4$ $(x +\frac 1 x)^2 = (4)^2$ $x^2 +\frac {1}{x^2} = 16 - 2 = 14$ $(x^2 +\frac {1}{x^2})^3 = (14)^3$ $x^6 +\frac {1}{x^6} + 3x^2\times\frac {1}{x^2}\times (x^2 + (\frac {1}{x^2}) = 2744$ $x^6 +\frac {1}{x^6} = 2744 - 42 = 2702$

Q.5: यदि x2 + 1 – 2x = 0, x > 0 है, तो x2(x– 2) का मान ज्ञात करें I
(A) 0
(B) -1
(C) 1
(D) $\sqrt2$
(SSC CHSL Aug 2021)

Ans : (B) -1
x2 + 1 – 2x = 0
x > 0
x = 1, so 1 + 1 – 2 = 0
Put x = 1 in x2(x– 2)
1 (1 – 2)
= -1

Q.6: यदि $x^2 -3\sqrt2x + 1 =0$ है , तो $x^3 + (\frac {1}{x^3})$ का मान ज्ञात करें I
(A) $15\sqrt6$
(B) $30\sqrt6$
(C) $45\sqrt2$
(D) $30\sqrt2$
(SSC CHSL Aug 2021)

Ans : (C) $45\sqrt2$ $x^2 - 3\sqrt2x + 1 = 0$ $x - 3\sqrt2 +\frac1x = 0$ $x +\frac 1x =3\sqrt2$ $(x +\frac 1x)^3 = (3\sqrt2)^3$ $x^3 +\frac {1}{x^3} + 3x\times\frac1x\times (x +\frac1x) = 54\sqrt2$ $x^3 +\frac{1}{x^3} = 54\sqrt2 - 9\sqrt2$ $= 45\sqrt2$

Q.7: यदि x – y = 4 और xy = 3 है , तो x3 – y3 का मान ज्ञात करें I
(A) 88
(B) 100
(C) 64
(D) 28
(SSC CHSL Aug 2021)

Ans : (B) 100
x – y = 4
(x – y)3 = 43
x3 – y3 – 3xy (x – y) = 64 [Put xy = 3, x – y =4]
x3 – y3 – 3 x 3 x 4 = 64
x3 – y3 = 64 + 36 = 100

Q.8: यदि $x -\frac 1x = 2\sqrt2$ है , तो $x^3 -\frac {1}{x^3}$ का मान ज्ञात करें I
(A) $12\sqrt2$
(B) $10\sqrt2$
(C) $20\sqrt2$
(D) $22\sqrt2$
(SSC CHSL Aug 2021)

Ans : (D) $22\sqrt2$ $x -\frac 1x =2\sqrt2$ $(x -\frac 1x)^3 = (2\sqrt2)^3$ $x^3 -\frac {1}{x^3} - 3\times x\times\frac 1x (x -\frac1x) = 16\sqrt2$ $x^3 -\frac {1}{x^3} - 3\times 2\sqrt2 = 16\sqrt2$ $x^3 -\frac {1}{x^3} = 16\sqrt2 + 6\sqrt2$ $= 22\sqrt2$

Q.9: यदि x + 2y = 19 और x3 + 8y3 = 361 है , तो xy का मान क्या होगा ?
(A) 58
(B) 56
(C) 55
(D) 57
(SSC CHSL Aug 2021)

Ans : (D) 57
x + 2y = 19
(x + 2y)3 = (19)3
x3 + 8y3 + 6xy (x + 2y) = 6859
x3 + 8y3 + 6xy x 19 = 6859
6xy x 19 = 6859 – 361 = 6498
xy = $\frac {6498}{19\times6}$
xy = 57

Q.10: यदि $(x^2 + \frac{1}{49x^2}) = 15\frac 57$ है , तो $(x +\frac{1}{7x})$ का मान क्या होगा ?
(A) 4
(B) $\pm 7$
(C) $\pm 4$
(D) 7
(SSC CHSL Aug 2021)

Ans : (C) $\pm 4$ $(x^2 + \frac{1}{49x^2}) = 15\frac57$ $(x^2 + \frac{1}{7x})^2 + 2\times x\times \frac {1}{7x} =\frac{110}{7} + \frac27$ $(x + \frac{1}{7x})^2 = \frac{112}{7} = 16$ $x + \frac{1}{7x} =\pm 4$

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