2x²-x-1=2x²-2x+x-1=2x(x-1)+(x-1)=(x-1)((2x+1)

and 4x²+8x+3.=4x²+6x+2x+3=2x(2x+3)+(2x+3)=(2x+3)(2x+1)

Here common factor=2x+1

hence HCF=2x+1

 
In  {tex}\triangle {/tex}OQP,DE || OQ
{tex}\frac{{PE}}{{EQ}} = \frac{{PD}}{{DO}}{/tex} .....(i)
In {tex}\triangle {/tex}OPR,DF {tex} \bot {/tex} OR
{tex}\frac{{PD}}{{DO}} = \frac{{PF}}{{FR}}{/tex} .....(ii)
From (i) and (ii) , we get
{tex}\frac{{PE}}{{EQ}} = \frac{{PF}}{{FR}}{/tex}
{tex}\therefore {/tex} From {tex}\triangle {/tex}PQR
EF {tex} \bot {/tex} OR

(2x-1) (x-3)=(x+5)

2x²-x-6x+3=x+5

2x²-8x-2=0

x²-4x-1=0

this is a quadratic equation

Given linear equation is
(3k + 1)x + 3 y - 2 = 0 .......... (i)
(k² + 1)x + (k - 2)y - 5 = 0 ............ (ii)
Compare with a₁x + b₁y + c = 0 and a₂x + b₂y and c₂ = 0
a₁ = 3k + 1 , b₁ = 3 , c₁ = -2
and a₂ = k²+ 1 , b₂ = k - 2, c₂ = -5
The given system of equations will have no solution, if
{tex} \frac { a_1 } { a_2 } = \frac { b_1 } { b_2} \neq \frac { c_1 } { c_2 }{/tex}
{tex} \frac { 3 k + 1 } { k ^ { 2 } + 1 } = \frac { 3 } { k - 2 } \neq \frac { - 2 } { - 5 }{/tex}
{tex}\Rightarrow \quad \frac { 3 k + 1 } { k ^ { 2 } + 1 } = \frac { 3 } { k - 2 } \text { and } \frac { 3 } { k - 2 } \neq \frac { 2 } { 5 }{/tex}
Now, {tex}\frac { 3 k + 1 } { k ^ { 2 } + 1 } = \frac { 3 } { k - 2 }{/tex}
{tex}\Rightarrow{/tex} (3k + 1)(k - 2)=3(k² + 1)
{tex}\Rightarrow{/tex} 3k² - 5k - 2 =3k² + 3
{tex}\Rightarrow{/tex} -5k - 2 =3
{tex}\Rightarrow{/tex} -5k = 5
{tex}\Rightarrow{/tex} k = -1
Clearly, {tex}\frac { 3 } { k - 2 } \neq \frac { 2 } { 5 }{/tex} for k = -1.
Hence, the given system of equations will have no solution for k = -1.

Let the polynomial be ax² + bx + c
and its zeroes be {tex}\alpha{/tex} and {tex}\beta{/tex}
Then, {tex}\alpha + \beta = \frac { 1 } { 4 } = - \frac { b } { a } \text { and } \alpha \beta = - 1 = \frac { c } { a }{/tex}
If a = 4, then b = -1 and c = -4
So, one quadratic polynomial which files
the given conditions is 4x² - x - 4
Or
If {tex}\alpha{/tex} and {tex}\beta{/tex} zeroes of the polynomials then standard form quadratic polynomial is given by
{tex}x ^ { 2 } - ( \alpha + \beta ) x + \alpha \beta{/tex}
Let {tex}\alpha = \frac { 1 } { 4 } \text { and } \beta = - 1{/tex}
Now,
{tex}= x ^ { 2 } - ( \alpha + \beta ) x + \alpha \beta{/tex}
{tex}= x ^ { 2 } - \left( \frac { 1 } { 4 } \right) x + ( - 1 ){/tex}
{tex}= \frac { 1 } { 4 } \left( 4 x ^ { 2 } - x - 4 \right){/tex}
Required polynomial is 4x² - x - 4

– 3

Solution: Here is a sample diagram related to the given question. 

               Given : {tex}\huge \angle A=\angle Q \space ; \space \space \space \angle B=\angle R{/tex}                            Thus , {tex}\huge \angle C=\angle P {/tex}

               Hence, {tex}\huge \triangle ABC\cong \triangle QRP {/tex}

               Now,  (i) is not possible and (ii), (iii), (iv) are possible conditions

Question: if {tex}\large \alpha \space and \space \beta{/tex} are the zeros of {tex}\large x^2 + 5 x + 5{/tex} then find the value of: (i)    {tex}(\alpha -1)+(\beta -1){/tex},(ii) {tex}\large \frac {\alpha} {\beta} +\frac { \beta} { \alpha}{/tex}

Solution: Roots of the equation    {tex}\huge x^2 + 5 x + 5 = {\frac {-5\pm \sqrt 5}{2}}{/tex}

                Thus,   {tex}\huge \alpha = {\frac {-5+ \sqrt 5}{2}} \space , \space \beta = {\frac {-5- \sqrt 5}{2}}{/tex}

                Now,  (i) {tex}\huge (\alpha -1)+(\beta -1) \space = \space (\alpha+\beta-2) \space = \space {\frac {-5+ \sqrt 5}{2}} + {\frac {-5- \sqrt 5}{2}} - 2{/tex} 

                         {tex}\huge \implies {\frac {-5+ \sqrt 5 -5-\sqrt 5}{2}} - 2{/tex}             {tex}\huge \implies {\frac {-10}{2}} - 2{/tex}

                         {tex}\huge \implies {\frac {-10-4}{2}} {/tex}          {tex}\huge \implies {\frac {-14}{2}} {/tex}       {tex}\huge \implies\boxed{ -7}{/tex}

                Now,  (i) {tex}\huge {\frac {\alpha}{\beta}}+{\frac {\beta} {\alpha}} \space = \frac{\alpha ^2 + \beta ^2}{\alpha \beta} {/tex} 

                         {tex}\large \implies Since, (i)\space The \space sum \space of \space zeroes, \space \alpha+\beta=5 ,(-b/a)\space \space \space ;\space \space \space (ii)\space The \space product \space of \space zeroes, \space \alpha\beta = 5, \space \space (c/a){/tex}  

Here is the international place value chart

Let the first terms are a and b

then their difference 100th terms=

=(a+99d)-(b+99d)=100

a-b=100

now difference 1000th terms

=a+999d-(b+999d)

=a-b=100

Buy hindi medium book

Given √2 is irrational number.
Let √2 = a / b wher a,b are integers b ≠ 0
we also suppose that a / b is written in the simplest form
Now √2 = a / b ⇒ 2 = a² / b² ⇒   2b² = a²
∴ 2b² is divisible by 2
⇒  a² is divisible by 2    
⇒  a is divisible by 2  
∴ let a = 2c
a² = 4c² ⇒ 2b² = 4c² ⇒ b² = 2c²
∴ 2c²  is divisible by 2
∴ b²  is divisible by 2
∴ b  is divisible by 2
∴a are b   are  divisible by 2 .
this contradicts our supposition that a/b is written in the simplest form
Hence our supposition is wrong
∴ √2 is irrational number.

This is Q.4 of Exercise 13.1, you have to understand the language of the question and solve it properly.
Its answer is 7cm for the greatest diameter and the Total Surface Area of the solid will be 332.5 cm².

The base area of the hemisphere is subtracted because it is hidden and is not visible from outside so we cant cover it.

Oh my god !! seems like you haven't practiced successive division. It is soo easy.
{tex}\huge 520 = 92\times5+50\\\space\space \huge 92= 50\times 1 + 42\\\space\space \huge 50=42\times1+8\\\space\space \huge 42=\space8\times5+2\\\space\space \space\space\huge 8= \space2\times4+0{/tex}

Here we ended up at 8, soo its divisor will be our HCF i.e. {tex}\large\boxed2{/tex}
 

5/x - 3/y = 2
(5y - 3x)/xy = 2
5y - 3x = 2xy

Prime factorization:
{tex}144 = 2 ^ { 4 } \times 3 ^ { 2 }{/tex}
{tex}198 = 2 \times 3 ^ { 2 } \times 11{/tex}
HCF = product of smallest power of each common prime factor in the numbers {tex}= 2 \times 3 ^ { 2 } = 18{/tex}

This is Q.13 of Exercise 10.2,

Suppose {tex}\sqrt [ 3 ] { 6 }{/tex} be rational number and {tex}\sqrt [ 3 ] { 6 } = \frac { a } { b }{/tex} where {tex}a\ and\ b{/tex} are co-prime and {tex}b\ne0{/tex}
{tex}\Rightarrow ( \sqrt [ 3 ] { 6 } ) ^ { 3 } = \frac { a ^ { 3 } } { b ^ { 3 } }{/tex}

{tex}\Rightarrow 6 = \frac { a ^ { 3 } } { b ^ { 3 } } {/tex}

{tex}\Rightarrow 6 . b ^ { 3 } = a ^ { 3 }{/tex}
{tex}\Rightarrow {/tex} {tex}a^3{/tex} is divisible by {tex}6{/tex} {tex}\Rightarrow{/tex} {tex}a{/tex} is divisible by {tex}6{/tex}.
Let {tex}a = 6c{/tex}
{tex}6b^3 = (6c)^3{/tex}
{tex}\Rightarrow \quad b ^ { 3 } = 36 c ^ { 3 }{/tex}
{tex}\Rightarrow {/tex} {tex}b^3{/tex} is divisible by {tex}6{/tex} {tex}\Rightarrow {/tex} {tex}b{/tex} is  divisible by {tex}6{/tex}.
{tex}\Rightarrow {/tex} {tex}a\ and\ b{/tex} have a common factor i.e, {tex}6{/tex}
{tex}\Rightarrow {/tex} {tex}a\ and\ b{/tex} are not co-prime which is a contradiction
{tex}\therefore \sqrt [ 3 ] { 6 }{/tex} is an irratonal.

f(x)=x²-3x-2

an b are zeros 

so a+b=3,ab=-2

now if the zeros are 1/2a+b and 1/2b+a

{tex}\begin{array}{l}\frac1{2a+b}+\frac1{2b+a}=\frac{2a+b+2b+a}{(2a+b)(2b+a)}=\frac{3(a+b)}{(2a+b)(2b+a)}\\=\frac{3(3)}{4ab+ab+2a^2+2b^2}=\frac9{5\ast(-2)+2\lbrack\;(a+b)^2-2ab)}\\=\frac9{-10+2\ast(9-2(-2)\rbrack}=\frac9{-10+2\ast(9+4)}=\frac9{-10+26}=\frac9{16}\\\\\end{array}{/tex}

{tex}\begin{array}{l}Product\;of\;zeros=\frac1{2a+b}\times\frac1{2b+a}=\frac1{(2a+b)(2b+a)}=\frac1{(2a+b)(2b+a)}\\=\frac1{4ab+ab+2a^2+2b^2}=\frac1{5\ast(-2)+2\lbrack\;(a+b)^2-2ab)}\\=\frac1{-10+2\ast(9-2(-2)\rbrack}=\frac1{-10+2\ast(9+4)}=\frac1{-10+26}=\frac1{16}\\\\\end{array}{/tex}

hence the polynomial

=x²-(9/16)x+1/16

or 16x²-9x+1

Hahaha, oh my god!!!, don't you think this is too much, you are not studying properly
{tex}\large p(x):x^3 -x^2+6x-5\\ putting, \space x=2\\ p(2):(2)^3-(2)^2+6\times2-5\\ p(2):8-4+12-5\\ p(2):\boxed {11}{/tex}

You just only have to replace '{tex}\large x {/tex}'  by the value given and perform the arithmetic operations.
Now, after seeing this, don't forget it.

And please write your qeustions properly while asking.

Let {tex} \alpha{/tex} and {tex} \frac { 1 } { \alpha }{/tex} be the zeros of (a² + 9)x² + 13x + 6a.
Then, we have
{tex} \alpha \times \frac { 1 } { \alpha } = \frac { 6 a } { a ^ { 2 } + 9 }{/tex}
⇒ 1 = {tex} \frac { 6 a } { a ^ { 2 } + 9 }{/tex}
⇒ a² + 9 = 6a
⇒ a² - 6a + 9 = 0
⇒ a² - 3a - 3a + 9 = 0
⇒ a(a - 3)  - 3(a - 3) = 0
⇒ (a - 3) (a - 3) = 0
⇒ (a - 3)² = 0
⇒ a - 3 = 0
⇒ a = 3
So, the value of a in given polynomial is 3.