Algebra

${ u }^{ 2 }-2u+2=0$
$\Longrightarrow \quad u=1\pm i$
So,$\alpha =1+i\quad and\quad \beta =1-i$
Now given that,$x=\cot { \theta  } -1$
so,$\displaystyle \frac { { (x+\alpha ) }^{ n }-{ (x+\beta ) }^{ n } }{ \alpha -\beta  } =\frac { { (\cot { \theta  } -1+1+i) }^{ n }-{ (\cot { \theta  } -1 }+1-i)^{ n } }{ 2i } \ \ $
$=\displaystyle \frac { { (\cot { \theta  } +i) }^{ n }-(\cot { \theta  } -i)^{ n } }{ 2i } =\frac { { (\cos { \theta  } +i\sin { \theta  } ) }^{ n }-{ (\cos { \theta  } -\sin { \theta  } ) }^{ n } }{ ({ \sin { \theta  }  })^{ n }(2i) } \ \ $
$=\displaystyle \frac { { e }^{ (in\theta ) }-{ e }^{ -(in\theta ) } }{ ({ \sin { \theta ) }  }^{ n }2i } \ \ $
=$\displaystyle \frac { (\cos { (n\theta ) } +i\sin { (n\theta )) } -(\cos { (n\theta ) } -i\sin { (n\theta )) }  }{ ({ \sin { \theta ) }  }^{ n }2i } =\frac { \sin { (n\theta ) }  }{ { (\sin { \theta ) }  }^{ n } } \ \ $

$cis\left( \theta  \right) =\cos { \theta  } +i\sin { \theta  } $
De Moivre's Theorem for fractional power:
${ \left( cis\theta  \right)  }^{ \frac { 1 }{ n }  }=cis\left( \frac { 2k\Pi +\theta  }{ n }  \right) $

${ \left( x-3 \right)  }^{ 3 }+1=0$
$\Longrightarrow x=3+{ \left( cis\left( \Pi  \right)  \right)  }^{ \frac { 1 }{ 3 }  }$
$x=3+{ \left( cis\left( \frac { 2k\Pi +\Pi  }{ 3 }  \right)  \right)  }$      ...{De Moivre's Theorem}
Where, $k=0,1,2$
for  $k=0$,
$x _{ 1 }=3+cis\left( \frac { \Pi  }{ 3 }  \right)$ 

for $k=1$,
$x _{ 2 }=3+cis\left( \Pi  \right) $

for $k=2,$
$x _{ 3 }=3+cis\left( \frac { 5\Pi  }{ 3 }  \right) $

$\Longrightarrow { x } _{ 1 }+{ x } _{ 3 }=6+cis\left( \frac { \Pi  }{ 3 }  \right) +cis\left( \frac { 5\Pi  }{ 3 }  \right) \ \Longrightarrow { x } _{ 1 }+{ x } _{ 3 }=7$
 
Ans: D

We have,

Given equation is

$\left( a-b \right){{x}^{2}}+\left( c-a \right)x+\left( b-c \right)=0$

On comparing that,

$A{{x}^{2}}+Bx+C=0$

Now,

$ A=\left( a-b \right) $

$ B=\left( c-a \right) $

$ C=\left( b-c \right) $

Roots are equal

Then,

$ D=0 $

$ {{B}^{2}}-4AC=0 $

$ \Rightarrow {{\left( c-a \right)}^{2}}-4\left( a-b \right)\left( b-c \right)=0 $

$ \Rightarrow {{c}^{2}}+{{a}^{2}}-2ac=4\left( ab-ac-{{b}^{2}}+bc \right) $

$ \Rightarrow {{c}^{2}}+{{a}^{2}}-2ac=4ab-4ac-4{{b}^{2}}+4bc $

$ \Rightarrow {{c}^{2}}+{{a}^{2}}-2ac+4ac=4ab-4{{b}^{2}}+4bc $

$ \Rightarrow {{\left( c+a \right)}^{2}}=4b\left( a-b+c \right) $

Hence, this is the answer

Let the common ratio of G.P is $r$ Therefore $\quad \beta =\alpha t,\alpha { t }^{ 2 }$
Equation $\alpha { x }^{ 2 }+2\alpha rx+\alpha { t }=0\quad 
\Rightarrow { x }^{ 2 }+2rx+{ t }^{ 2 }=0....(i)$
Given equation (i) and ${ x }^{ 2 }+x-1=0....(ii)$ has a common root
$(i)-(ii)\Rightarrow (2e-1)x+({ r }^{ 2 }+1)=0\Rightarrow x=\cfrac { -\left( { r }^{ 2 }+1 \right)  }{ 2r-1 } ....(iii)\quad $
Putting (iii) in equation (ii) $\Rightarrow { \left( { r }^{ 2 }+1 \right)  }^{ 2 }-\left( { r }^{ 2 }+1 \right) (2r-1)-{ \left( { 2r }^{ 2 }-1 \right)  }^{ 2 }=0\Rightarrow { r }^{ 4 }-2{ r }^{ 3 }-{ r }^{ 2 }+2r+1=0....(iv)$
dividing equation (iv) by ${r}^{2}$ $\Rightarrow { \left( r-\cfrac { 1 }{ r }  \right)  }^{ 2 }-2{ \left( r-\cfrac { 1 }{ r }  \right)  }+1=0\Rightarrow { \left( r-\cfrac { 1 }{ r } -1 \right)  }^{ 2 }=0\Rightarrow \cfrac { r-1 }{ r } =1....(v)\quad $
$\left( \gamma -\alpha  \right) \beta =\left( \alpha { r }^{ 2 }-\alpha  \right) \times \alpha r={ \alpha  }^{ 2 }\left( { \alpha  }^{ 2 }-1 \right) r={ \alpha  }^{ 2 }(r-1)={ \alpha  }^{ 2 }{ r }^{ 2 }$
(using $(v)=\alpha \times \alpha { t }^{ 2 }\quad $

The quadratic formula is a formula for solving quadratic equations. It is named after al-Khwarizmi, who derived it in the 9th century.

The quadratic formula has a wide range of applications, including solving quadratic equations, finding the roots of polynomials, and factoring polynomials.

Bhaskara II's formula for solving quadratic equations is given by $$x = (-b ± √(b^2 - 4ac)) / 2a$$. This formula is still used today to solve quadratic equations.

Add 5 to both sides of the equation: (3x - 5 + 5 = 10 + 5). Simplify: (3x = 15). Divide both sides by 3: (x = 15/3). Simplify: (x = 5).

Distribute the 2: (2x + 6 = 10). Subtract 6 from both sides: (2x + 6 - 6 = 10 - 6). Simplify: (2x = 4). Divide both sides by 2: (2x/2 = 4/2). Simplify: (x = 2).

Distribute the 4: (8x - 4 = 20). Add 4 to both sides: (8x - 4 + 4 = 20 + 4). Simplify: (8x = 24). Divide both sides by 8: (8x/8 = 24/8). Simplify: (x = 3).

Distribute the 3: (3x - 6 = 15). Add 6 to both sides: (3x - 6 + 6 = 15 + 6). Simplify: (3x = 21). Divide both sides by 3: (3x/3 = 21/3). Simplify: (x = 7).

Brahmagupta's formula for solving a quadratic equation is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Bhaskara I's formula for solving a quadratic equation is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

To solve the system of equations, we can use the substitution method. First, solve one of the equations for one of the variables. For example, we can solve the first equation for (x): (x = 5 - 2y). Then, substitute this expression for (x) into the other equation: (2(5 - 2y) - y = 1). This gives us (10 - 4y - y = 1), which simplifies to (-5y = -9). Dividing both sides by (-5), we get (y = 9/5). Substituting this value of (y) back into the first equation, we get (x + 2(9/5) = 5), which simplifies to (x = 1). Therefore, the solution to the system of equations is ((x, y) = (1, 2)).

To solve the system of equations, we can use the addition method. First, add the two equations together to get (2x = 6). Dividing both sides by 2, we get (x = 3). Substituting this value of (x) back into the first equation, we get (3 + y = 5), which simplifies to (y = 2). Therefore, the solution to the system of equations is ((x, y) = (3, 2)).