use quadratic formula to solve :

x2+2ax+(a 2 -b2)=0

x2 - 2ax + (a2 - b2) = 0

x2 - 2ax + (a + b)(a - b) = 0

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

x(x - a + b) - (a + b)(x - a + b) = 0

(x - a + b)(x - a - b) = 0

x = a + b , x = a - b

  • 0

when we are asked to use quad. formula method

we let A=coefficient of x2 = 1 , B=coefficient of x = 2a , C= constant term = a2-b2

now D = B2 - 4AC

  = ( 2a )2 - 4 x 1 x ( a2 - b2 )

  = 4a2 - 4 ( a2 - b2 )

  = 4a2 - 4a2+ 4b2

   = 4b2

since , the square of any integer cant be negative

so D > 0

now ... first root = - B2 +root D / 2A

  = - ( 2a )2 + root 4b2 / 2 x 1

  =  - 4a2 + 2b / 2

  = -2 ( 2a2 - b ) / 2

  =  -2a2 + b

second root= -B2 - rootD / 2A

  = - ( 2a )2 - root 4b2 / 2 x 1

  =  - 4a2 - 2b / 2

  =  -2 ( 2a2 + b) / 2

  =  - 2a2 - b

  • 3
a+b,a+b
  • -4
a+b,a-b
  • -1
Hi Answer is in image Thank you

  • 20
What are you looking for?