Question

If 1, a1, a2, …, an−1 be the roots of the equation x

n - 1 = 0, n ∈, n ≥ 2 show that

n = (1 − a1) (l − a2)(1 − a3) … (1 − an-1)​

Answer 1

Answer:

Solution :

\bf{\red{\underline{\bf{Given\::}}}}

Given:

The denominator of a fraction is 1 more than double of the numerator.If both the numerator and the denominator are increased by 4,the fraction becomes 3/5.

\bf{\red{\underline{\bf{To\:find\::}}}}

Tofind:

The fraction.

\bf{\red{\underline{\bf{Explanation\::}}}}

Explanation:

Let the numerator be r

Let the denominator be m

A/q

\longrightarrow\sf{m=2r+1...................(1)}⟶m=2r+1...................(1)

&

\begin{lgathered}\longrightarrow\sf{\dfrac{r+4}{m+4} =\dfrac{3}{5} }\\\\\\\longrightarrow\sf{5(r+4)=3(m+4)}\\\\\\\longrightarrow\sf{5r+20=3m+12}\\\\\\\longrightarrow\sf{5r-3m=12-20}\\\\\\\longrightarrow\sf{5r-3m=-8}\\\\\\\longrightarrow\sf{5r-3(2r+1)=-8\:\:\:[from(1)]}\\\\\\\longrightarrow\sf{5r-6r-3=-8}\\\\\\\longrightarrow\sf{-r=-8+3}\\\\\\\longrightarrow\sf{\cancel{-}r=\cancel{-}5}\\\\\\\longrightarrow\sf{\blue{r=5}}\end{lgathered}

⟶

m+4

r+4

=

5

3

⟶5(r+4)=3(m+4)

⟶5r+20=3m+12

⟶5r−3m=12−20

⟶5r−3m=−8

⟶5r−3(2r+1)=−8[from(1)]

⟶5r−6r−3=−8

⟶−r=−8+3

⟶

−

r=

−

5

⟶r=5

Putting the value of r in equation (1),we get;

\begin{lgathered}\longrightarrow\sf{m=2(5)+1}\\\\\longrightarrow\sf{m=10+1}\\\\\longrightarrow\sf{\blue{m=11}}\end{lgathered}

⟶m=2(5)+1

⟶m=10+1

⟶m=11

Thus;

Step-by-step explanation:

Solution :

\bf{\red{\underline{\bf{Given\::}}}}

Given:

The denominator of a fraction is 1 more than double of the numerator.If both the numerator and the denominator are increased by 4,the fraction becomes 3/5.

\bf{\red{\underline{\bf{To\:find\::}}}}

Tofind:

The fraction.

\bf{\red{\underline{\bf{Explanation\::}}}}

Explanation:

Let the numerator be r

Let the denominator be m

A/q

\longrightarrow\sf{m=2r+1...................(1)}⟶m=2r+1...................(1)

&

\begin{lgathered}\longrightarrow\sf{\dfrac{r+4}{m+4} =\dfrac{3}{5} }\\\\\\\longrightarrow\sf{5(r+4)=3(m+4)}\\\\\\\longrightarrow\sf{5r+20=3m+12}\\\\\\\longrightarrow\sf{5r-3m=12-20}\\\\\\\longrightarrow\sf{5r-3m=-8}\\\\\\\longrightarrow\sf{5r-3(2r+1)=-8\:\:\:[from(1)]}\\\\\\\longrightarrow\sf{5r-6r-3=-8}\\\\\\\longrightarrow\sf{-r=-8+3}\\\\\\\longrightarrow\sf{\cancel{-}r=\cancel{-}5}\\\\\\\longrightarrow\sf{\blue{r=5}}\end{lgathered}

⟶

m+4

r+4

=

5

3

⟶5(r+4)=3(m+4)

⟶5r+20=3m+12

⟶5r−3m=12−20

⟶5r−3m=−8

⟶5r−3(2r+1)=−8[from(1)]

⟶5r−6r−3=−8

⟶−r=−8+3

⟶

−

r=

−

5

⟶r=5

Putting the value of r in equation (1),we get;

\begin{lgathered}\longrightarrow\sf{m=2(5)+1}\\\\\longrightarrow\sf{m=10+1}\\\\\longrightarrow\sf{\blue{m=11}}\end{lgathered}

⟶m=2(5)+1

⟶m=10+1

⟶m=11

Thus;