Question

Find the solution of the equation
x2 + x -(a+1)(a +2)=0.​

Answer 1

Answer:

x²+(a+2)x-(a+1)x-(a+2)(a+1)=0

x(x+a+2)-(a+1){x+a+2}=0

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

x=-(a+2),a+1

Answer 2

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x² + x - ( a + 1 ) ( a + 2 )

a = 1 , b = 1 , c = ( a + 1 )( a + 2 )

Finding Determinant ,

D = b² - 4ac

D = 1² - 4 (1) ( a + 1 ) ( a + 2 )

D = 1 - 4 x a ( a + 2 ) + 1 ( a + 2 )

D = 1 - 4 x a² + 2a + a + 2

D = 1 - 4a² + 3a + 2

D = 3 - 4a² + 3a

D = 4a² - 3a - 3

[ Multiplying by (-1) and rearranging terms ]

x = b² - √4ac / 2a

x = 1² - √4a² - 3a - 3 / 2 x 1

x = 1 - √4a² - 3a - 3 / 2

Therefore ,

x = 1 - √4a² - 3a -3 / 2 or x = - 1 - √4a² - 3a - 3

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