a ) G i v e n t h a t f ( x ) = a x 2 + b x − 2 f ( 1 ) = 7 a n d f ( 2 ) = 10 S u b s t i t u t i n g x = 1 i n f ( x ) a n d e q u a t i n g t o 7 , w e g e t a ( 1 ) 2 + b ( 1 ) − 2 = 7 ⇒ a + b = 9... ( 1 ) a) \ Given \ that \ f(x)=ax^2+bx-2 \\ f(1)=7 \ and \ f(2)=10\\
Substituting \ {x= 1} \ in \ f(x) \ and \ equating \ to \ 7, \ we \ get \\
a(1)^2+b(1)-2=7 \\\Rightarrow \ a+b =9 ...(1) \\ a ) G i v e n t ha t f ( x ) = a x 2 + b x − 2 f ( 1 ) = 7 an d f ( 2 ) = 10 S u b s t i t u t in g x = 1 in f ( x ) an d e q u a t in g t o 7 , w e g e t a ( 1 ) 2 + b ( 1 ) − 2 = 7 ⇒ a + b = 9... ( 1 ) S u b s t i t u t i n g x = 2 i n f ( x ) a n d e q u a t i n g t o 10 , w e g e t a ( 2 ) 2 + b ( 2 ) − 2 = 10 4 a + 2 b = 12 D i v i d i n g w i t h 2 , w e g e t 2 a + b = 6... ( 2 ) Substituting \ {x = 2} \ in \ f(x) \ and \ equating \ to \ 10, \ we \ get \\ a(2)^2+b(2)-2=10 \\ 4a+2b=12 \\ Dividing \ with \ {2}, \ we \ get \ 2a+b=6 ...(2)\\ S u b s t i t u t in g x = 2 in f ( x ) an d e q u a t in g t o 10 , w e g e t a ( 2 ) 2 + b ( 2 ) − 2 = 10 4 a + 2 b = 12 D i v i d in g w i t h 2 , w e g e t 2 a + b = 6... ( 2 ) F r o m t h e e q u a t i o n ( 1 ) , b = 9 − a . . . ( 3 ) S u b s t i t u t i n g b = 9 − a i n ( 2 ) , w e g e t 2 a + 9 − a = 6 ⇒ a + 9 = 6 ⇒ a = 6 − 9 = − 3 ∴ a = − 3 S u b s t i t u t i n g a = − 3 , i n ( 3 ) , w e g e t b = 9 − ( − 3 ) = 12 ∴ b = 12. From \ the \ equation (1), \ { b = 9-a} ...(3)\\
Substituting \ {b=9-a} \ in \ {(2)},\ we \ get \\ 2a+9-a=6 \\ \Rightarrow \ {a+9=6} \\
\Rightarrow \ {a= 6-9=-3}\\
\therefore a =-3\\
Substituting \ {a=-3}, in \ (3), we \ get \ {b= 9-(-3)=12} \\
\therefore b=12. F ro m t h e e q u a t i o n ( 1 ) , b = 9 − a ... ( 3 ) S u b s t i t u t in g b = 9 − a in ( 2 ) , w e g e t 2 a + 9 − a = 6 ⇒ a + 9 = 6 ⇒ a = 6 − 9 = − 3 ∴ a = − 3 S u b s t i t u t in g a = − 3 , in ( 3 ) , w e g e t b = 9 − ( − 3 ) = 12 ∴ b = 12. b ) G i v e n t h a t f ( x ) = g ( x ) − 1 a n d g ( x ) = 3 x + 2 ⇒ a x 2 + b x − 2 = 3 x + 2 S u b s t i t u t i n g a = − 3 a n d b = 12 , w e g e t b)\ Given \ that \ {f(x)= g(x)-1} \ and \ {g(x)=3x+2} \\
\Rightarrow \ ax^2+bx-2=3x+2\\
Substituting \ {a= -3} \ and \ {b=12} , \ we \ get \\ b ) G i v e n t ha t f ( x ) = g ( x ) − 1 an d g ( x ) = 3 x + 2 ⇒ a x 2 + b x − 2 = 3 x + 2 S u b s t i t u t in g a = − 3 an d b = 12 , w e g e t − 3 x 2 + 12 x − 2 = 3 x + 2 − 1 ⇒ − 3 x 2 + 12 x − 3 x − 2 − 2 + 1 = 0 ⇒ − 3 x 2 + 9 x − 3 = 0 D i v i d i n g w i t h − 3 , w e g e t x 2 − 3 x + 1 = 0 T h e r o o t s o f a q u a d r a t i c a r e g i v e n b y x = − b ± b 2 − 4 a c 2 a -3x^2+12x-2=3x+2-1\\
\Rightarrow \ -3x^2+12x-3x-2-2+1=0\\
\Rightarrow \ -3x^2+9x-3=0 \\
Dividing \ with \ {-3},\ we \ get
\ {x^2-3x+1=0} \\
The \ roots \ of \ a \ quadratic \ are \ given \ by \\
x= \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ − 3 x 2 + 12 x − 2 = 3 x + 2 − 1 ⇒ − 3 x 2 + 12 x − 3 x − 2 − 2 + 1 = 0 ⇒ − 3 x 2 + 9 x − 3 = 0 D i v i d in g w i t h − 3 , w e g e t x 2 − 3 x + 1 = 0 T h e roo t s o f a q u a d r a t i c a re g i v e n b y x = 2 a − b ± b 2 − 4 a c S u b s t i t u t i n g a = 1 , b = − 3 a n d c = 1 , w e g e t x = − ( − 3 ) ± ( − 3 ) 2 − 4 ( 1 ) ( 1 ) 2 ( 1 ) = 3 ± 9 − 4 2 = 3 ± 5 2 Substituting \ {a=1},\ {b=-3} \ and \ {c=1}, \ we \ get \\
x=\frac{-(-3)\pm\sqrt{(-3)^2-4(1)(1)}}{2(1)}\\
=\frac{3\pm\sqrt{9-4}}{2}\\
=\frac{3\pm\sqrt{5}}{2}\\ S u b s t i t u t in g a = 1 , b = − 3 an d c = 1 , w e g e t x = 2 ( 1 ) − ( − 3 ) ± ( − 3 ) 2 − 4 ( 1 ) ( 1 ) = 2 3 ± 9 − 4 = 2 3 ± 5 ∴ x 1 = 3 + 5 2 a n d x 2 = 3 − 5 2 . \therefore { x_{1}=\frac{3+\sqrt{5}}{2}} \ and \ {x_{2}= \frac{3-\sqrt{5}}{2}}. ∴ x 1 = 2 3 + 5 an d x 2 = 2 3 − 5 .
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