Solve for x :
9x2 - 9( a + b )x + ( 2a2 + 5ab + 2b2 )
Solve this by both factorisation and by using quadratic formula.
1.) Using factorization
9x2 -9( a + b ) x + (2a2 + 5ab + 2 b2) = 0
⇒9x2 -9( a + b )x + ( 2a+b ) ( a+2b) =0 (By middle term factorization of 2a2 + 5ab + 2b2=0)
⇒9x2 - ( 6a+3b )x - ( 3a+ 6b ) x + ( 2a+b ) ( a+2b) =0
⇒3x ( 3 x - (2a+b )) - ( a + 2 b ) ( 3x -( 2a+b) ) =0
⇒( 3x -(2a +b) ) ( 3x -(a+2b) ) =0
2.) Using the quadratic equation
9x 2 – 9(a + b) x + (2a 2 + 2b 2 + 5ab) = 0
Comparing this equation with the standard equation: Ax 2 + Bx + C = 0
Here, A = 9, B = –9 (a + b), C = (2a 2 +2b 2 + 5ab)
∴ Discriminant, D = B2 – 4AC
⇒ D = {–9 (a + b)}2 – 4 � 9 � (2a 2 + 2b 2 + 5ab)
⇒ D = 81(a + b)2 – 36 (2a 2 + 2b 2 + 5ab)
⇒ D = 81(a 2 + b 2+ 2ab) – 72a 2 – 72b 2 – 180ab
⇒ D = 9a 2 + 9b 2 – 18ab
⇒ D = 9(a 2 – 2ab + b 2)
⇒ D = 9(a – b)2 ≥ 0
As D ≥ 0, therefore, the roots of equation are real.
Cheers!!!
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