Question

If α and β are the roots of ax²+bx+c=0, then the equation whose roots are 1/aα+b and 1/aβ+b is:-
A) a cx² - bx + 1 = 0
B) a cx² + bx - 1 = 0
C) a cx² - bx - 1 = 0
D) bx² - ax + 1 = 0

Answer 1

Step-by-step explanation:

ANSWER

ax

2

+bx+c=0

α+β=

a

−b

αβ=

a

c

(i) equation whose roots are

α

2

1

,

β

2

1

Sum of roots =

α

2

1

+

β

2

1

=

(αβ)

2

α

2

+β

2

=

(αβ)

2

(α+β)

2

−2αβ)

=

a

2

c

2

a

2

b

2

−

a

2c

=

c

2

b

2

−2ac

\

product =

α

2

1

β

2

1

=

c

2

a

2

∴ equation is

c

2

x

2

+(2ac−b

2

)x+a

2

=0

(ii)

aα+β

1

,

aβ+α

1

Sum ⇒

a

2

αβ+α

2

a+aβ

2

+αβ

aβ+aα+β

⇒

(a

2

+1)(αβ)+a(αβ)

(α+β)(1+a)

⇒

(a

2

+1)

a

c

+a(

a

2

b

2

−

a

2c

)

(1+a)(

a

b

)

⇒

(a

2

+1)c+a(b

2

−2ac)

−(1+a)b

⇒

c(1+a

2

)+ab

2

−(1+a)b

Product

c(1−a

2

)+ab

2

a

∴(c(1−a

2

)+ab

2

)x

2

+(1+a)bx+a=0

(iii) α+

β

1

,β+

α

1

Sum ⇒

β

αβ+1

+

α

αβ+1

⇒

αβ

α

2

β(αβ)+(α

2

+β

⇒

a

−b

+

c

−b

−b[

ac

a+c

]

Product

αβ

(αβ+1)

2

⇒

a

2

a

c

(c+a)

2

⇒(α+β)+

αβ

αβ

[

a+c

ac

]x

2

+bx+(a+c)=0

Answer 2

ANSWER:

(A) a cx² - bx + 1 = 0

STEP-BY-STEP EXPLANATION:

$\alpha +\beta =\frac{-b}{a} \\\alpha \beta = \frac{c}{a}\\\\\frac{1}{a\alpha +b}+\frac{1}{a\beta +b}= \frac{a\beta +b+a\alpha +b}{(a\alpha +b)(a\beta +b)}$

                     = $\frac{a(\alpha +\beta )+2b}{a^{2} \alpha \beta +ab\alpha +ab\beta +b^{2} }$

                     = $\frac{a(\alpha +\beta )+2b}{a^{2} \alpha \beta +ab(\alpha +\beta )+b^{2} }$

                     =  $\frac{a(\frac{-b}{a} )+2b}{a^{2} (\frac{c}{a}) +ab(\frac{-b}{a}) +b^{2} }$

                     = $\frac{b}{ac}$

$(\frac{1}{a\alpha +b} )(\frac{1}{a\beta +b} )=\frac{1}{a^{2} \alpha \beta +ab\alpha +ab\beta +b^{2} }$

                      = $\frac{1}{a^{2} \alpha \beta +ab(\alpha +\beta) +b^{2} }$

                      = $\frac{1}{a^{2} (\frac{c}{a}) +ab(\frac{-b}{a})+b^{2} }$

                      = $\frac{1}{ac}$

∴ New equation = $x^{2} -(\frac{1}{a\alpha +b}+\frac{1}{a\beta +b})x + (\frac{1}{a\alpha +b}*\frac{1}{a\alpha +b} )=0$

                            = $x^{2} - (\frac{b}{ac} )x+\frac{1}{ac} =0$

                            = $acx^{2} -bx+1=0$

                            = option (A)

Hope it helps.