Question

1. Each root of x^2- bx+c=0 is decreased by 2
resulting equation is x^2– 2x+1=0, then
(a) b = 6,c = 9
(b) b = 3,c = 5
(c) b=2,c=-1
(d) b = - 4,= 3​

Answer 1

Step-by-step explanation:

Let α,β are the roots of equation ax2+bc+c=0

Then, α−1 and β−1 are the roots of equation 2x2+8x+2=0

Then, sum of roots, (α−1)+(β−1)=−82

⇒α+β=−4+2

⇒α+β=−2→(1)

Product of roots, (α−1)(β−1)=22

⇒αβ−α−β+1=1

⇒αβ−(α+β)=0

⇒αβ=α+β

⇒αβ=−2→(2)

As, α and β are the roots of ax2+bx+c=0

∴−ba=α+βandca=αβ

⇒−ba=−2andca=−2

⇒b=2aandca=−2a

⇒b=−c

So, option 2 is the correct option.

Answer 2

Solution: Option (B)

Explanation:

let the zeroes be $\rm \alpha\: and\:\beta$ for the polynomial, $\rm\color{red} x^2- bx+c=0$

Comparing with the form Ax² + Bx + C = 0

A = 1

B = -b

C = c

Sum of the zeroes =$\frac{-B}{A}$

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

$\rm \alpha + \beta = b$ ---> (1)

Product of the zeroes =$\frac{C}{A}$

$\rm \alpha \beta = \frac{c}{1}$

$\rm \alpha \beta = c$ ---> (2)

If the zeroes are decreased by 2, then the zeroes will be $\rm (\alpha - 2) \:and\:( \beta - 2)$respectively for the resulting polynomial, $\rm\color{red} x^2- 2x+1=0$

Comparing with the form ax² + bx + c = 0

a = 1

b = -2

c = 1

Sum of the zeroes =$\frac{-b}{a}$

$\rm (\alpha - 2) + (\beta - 2) = \frac{-(-2)}{1}$

$\rm \alpha + \beta - 4 = 2$

$\rm \alpha + \beta = 6$

$\rm \fbox{ b = 6 }$ [from equation (1) ]

Product of the zeroes =$\frac{c}{a}$

$\rm (\alpha - 2) (\beta - 2) = \frac{1}{1}$

$\rm \alpha \beta - 2\alpha - 2\beta + 4 = 1$

$\rm \alpha \beta - 2(\alpha + \beta) + 4 = 1$

$\rm \alpha \beta - 2(6) + 4 = 1$

$\rm \alpha \beta - 12 + 4 = 1$

$\rm \alpha \beta - 8 = 1$

$\rm \alpha \beta = 9$

$\rm\fbox{ c = 9 }$ [from equation (2) ]