Question

100. The roots of x² - bx+c = 0 are each decreased by 2. The resulting equation is x² - 2x+1=0. Then​

Answer 1

$\Huge{\underline{\underline{\mathfrak{Correct \ Question \colon}}}}$

Each root of the equation x² - bx + c = 0 is decreased by 2 resulting in the equation x² - 2x + 1 = 0,find the values of b and c.

$\Huge{\underline{\underline{\mathfrak{Answer \colon}}}}$

Given Equation,

$\large{ \sf{x {}^{2} - bx + c = 0 }}..........(1)$

On reducing the roots of above equation by 2,the resulting equation is :

$\sf{x {}^{2} - 2x + 1 = 0 }$

To find

Values of b and c in the first equation

• Consider x² - 2x + 1 = 0

$\sf{x {}^{2} - 2x + 1 = 0 } \\ \\ \leadsto \: \sf{x {}^{2} - x - x + 1 = 0 } \\ \\ \leadsto \: \sf{x(x - 1) - 1(x - 1) = 0} \\ \\ \leadsto \: \sf{(x - 1) {}^{2} = 0 } \\ \\ \leadsto \: \boxed{\sf{x = 1}}$

We get 1 as the root of the second equation

》 3 would have been the root of the first equation

$\: \sf{x = 3} \\ \\ \rightarrow \: \sf{(x - 3) = 0}$

• Consider x² - bx + c = 0,

Product of its roots would give us the equation back.

$\sf{(x - 3)(x - 3) = 0} \\ \\ \implies \: \sf{x(x - 3) - 3(x - 3) = 0} \\ \\ \implies \: \sf{x {}^{2} - 3x - 3x + 9 = 0 } \\ \\ \implies \: \underline{ \boxed{\sf{x {}^{2} - 6x + 9 = 0 }}}................(2)$

Comparing equations (1) and (2),we get :

b = 6 and, c = 9

Answer 2

Answer:

Each root of the equation x² - bx + c = 0 is decreased by 2 resulting in the equation x² - 2x + 1 = 0,find the values of b and c.

Step-by-step explanation:

\Huge{\underline{\underline{\mathfrak{Answer \colon}}}}

Answer:

Given Equation,

\large{ \sf{x {}^{2} - bx + c = 0 }}..........(1)x

2

−bx+c=0..........(1)

On reducing the roots of above equation by 2,the resulting equation is :

\sf{x {}^{2} - 2x + 1 = 0 }x

2

−2x+1=0

To find

Values of b and c in the first equation

Consider x² - 2x + 1 = 0

\begin{gathered}\sf{x {}^{2} - 2x + 1 = 0 } \\ \\ \leadsto \: \sf{x {}^{2} - x - x + 1 = 0 } \\ \\ \leadsto \: \sf{x(x - 1) - 1(x - 1) = 0} \\ \\ \leadsto \: \sf{(x - 1) {}^{2} = 0 } \\ \\ \leadsto \: \boxed{\sf{x = 1}}\end{gathered}

x

2

−2x+1=0

⇝x

2

−x−x+1=0

⇝x(x−1)−1(x−1)=0

⇝(x−1)

2

=0

⇝

x=1

We get 1 as the root of the second equation

》 3 would have been the root of the first equation

\begin{gathered}\: \sf{x = 3} \\ \\ arrow \: \sf{(x - 3) = 0}\end{gathered}

x=3

arrow(x−3)=0

Consider x² - bx + c = 0,

Product of its roots would give us the equation back.

\begin{gathered}\sf{(x - 3)(x - 3) = 0} \\ \\ \implies \: \sf{x(x - 3) - 3(x - 3) = 0} \\ \\ \implies \: \sf{x {}^{2} - 3x - 3x + 9 = 0 } \\ \\ \implies \: \underline{ \boxed{\sf{x {}^{2} - 6x + 9 = 0 }}}................(2)\end{gathered}

(x−3)(x−3)=0

⟹x(x−3)−3(x−3)=0

⟹x

2

−3x−3x+9=0

⟹

x

2

−6x+9=0

................(2)

Comparing equations (1) and (2),we get :

b = 6 and, c = 9