Question

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


${x}^{2} - (2a + 3)x + {a}^{2} + 3a + 2 = 0$
Solve both quadratic equations ;-

Answer 1

Hey Mate !

Here is your solution :

1.

p( x ) = ( a - b )x² + ( b - c )x + ( c - a ) = 0

Here,

Coefficient of x² = ( a - b )

Coefficient of x = ( b - c )

Constant term = ( c - a )

Now,

Factors of constant term = ±1 , ±( c - a )

When, x = 1


=> p( x ) = ( a - b )x² + ( b - c )x + ( c -a )

=> p( 1 ) = ( a - b )1² + ( b - c )1 + ( c - a )

=> p( 1 ) = ( a - b ) + ( b - c ) + ( c - a )

=> p( 1 ) = a - b + b - c + c - a

•°• p( 1 ) = 0

Hence,

By factor theorem,

1 is a root of this equation.

Let another zero is α.

Now,

=> Sum of zeroes = - Coefficient of x / Coefficient of x²


=> 1 + α = -( b - c ) / ( a - b )

=> 1 + α = ( c - b ) / ( a - b )

=> α = [ ( c - b ) / ( a - b ) ] - 1

=> α = [ ( c - b ) - ( a - b ) ] / ( a - b )

=> α = [ c - b - a + b ] / ( a - b )

=> α = ( c - a ) / ( a - b )

Therefore,

Its zeroes are 1 and ( c - a ) / ( a - b ).


2.

=> x² - ( 2a + 3 )x + a² + 3a + 2 = 0

=> x² - ( 2a + 3 )x + ( a² + 3a + 2 ) = 0

=> x² - ( 2a + 3 )x + ( a² + 2a + a + 2 ) = 0

=> x² - ( 2a + 3 )x + [ a(a+ 2 ) + 1( a + 2 )] =0

=> x² - ( 2a + 3 )x + ( a + 2 ) ( a + 1 ) = 0

Splitting the middle term,

=> x² - [ ( a + 2 ) + ( a + 1 ) ]x + ( a + 2 ) ( a + 1 ) = 0

=> x² - ( a + 2 )x - ( a + 1 )x + ( a + 2 ) ( a + 1 ) = 0


=> x( x - a - 2 ) - ( a + 1 ) [ x - ( a + 2 ) ] = 0

=> x( x - a - 2 ) - ( a + 1 ) ( x - a - 2 ) = 0

Taking out ( x - a - 2 ) as common,

=> ( x - a - 2 ) ( x - a - 1 ) = 0

=> ( x - a - 2 ) = 0 ÷ ( x - a - 1 )

=> x - a - 2 = 0

•°• x = a + 2

" Or "

=> ( x - a - 2 ) ( x - a - 1 ) = 0

=> ( x - a - 1 ) = 0 ÷ ( x - a - 2 )

=> ( x - a - 1 ) = 0

•°• x = a + 1

Therefore,

Zeroes are ( a + 1 ) , ( a + 2 ).

Answer 2

$\large{\mathbf{HELLO \: \: DEAR,}}$

$\mathbf{(a - b)x^2 + (b - c)x + (c - a) = 0}\\ \\ \mathbf{(a - b)x^2 - (a - b)x - (c - a)x + (c - a) = 0} \\ \\ \mathbf{(ax - bx)(x - 1) - (c - a)(x - 1) = 0} \\ \\ \mathbf{(X - 1)[(a - b)x - c + a)]=0} \\ \\ \mathbf{X = 1 , x = (c - a)/(a - b)}\\ \\ \\ \\ \: \: \: \large{\mathbf{SECOND \: \: \: QUESTION:-}} \\ \\ \\ \mathbf{X^2 - (2a + 3)x + a^2 + 3a + 2 = 0} \\ \\ \mathbf{X^2 - (2a + 3)x + a^2 + 2a + a + 2 = 0} \\ \\ \mathbf{X^2 - (2a + 3)x + a(a + 2) + (a + 2) = 0} \\ \\ \mathbf{X^2 - (2a + 3)x + (a + 1)(a + 2) = 0} \\ \\ \mathbf{X^2 - (a + a + 1 + 2)x + (a + 1)(a + 2) = 0} \\ \\ \mathbf{X^2 - (a + 1)x - (a + 2)x + (a + 1)(a + 2) = 0} \\ \\ \mathbf{x(x - a - 1) - (a + 2)(x - a - 1 ) = 0} \\ \\ \mathbf{(X - a - 2)(x - a - 1) = 0} \\ \\ \mathbf{X = a + 2 , \: \: x = a + 1}$


$\large{\mathbf{\underline{I \: \: HOPE \: \: ITS \: \: HELP \: \: YOU \: \: DEAR, \: \: THANKS}}}$