Question

if a = 1 + root 7 ,find the value of - 6/a

Answer 1

Given that,

$a=1+\sqrt{7}$

To find,

The value of $\dfrac{-6}{a}$.

Solution,

To find the value of $\dfrac{-6}{a}$, we put the value of a as :

$\dfrac{-6}{a}=\dfrac{-6}{1+\sqrt{7} }$

Rationalizing by multiplying and dividing by $1-\sqrt{7}$. So,

$\dfrac{-6}{a}=\dfrac{-6}{1+\sqrt{7} }\times \dfrac{1-\sqrt7}{1-\sqrt7}\\\\=\dfrac{-6(1-\sqrt7)}{1^2-(\sqrt{7})^2 }\\\\=\dfrac{-6+6\sqrt7}{1-7}\\\\=\dfrac{-6+6\sqrt7}{-6}\\\\=\dfrac{-6(1-\sqrt7)}{-6}\\\\=1-\sqrt7$

So, the value of $\dfrac{-6}{a}$ is equal to $1-\sqrt{7}$.

Answer 2

Step-by-step explanation:

Given that,

a=1+\sqrt{7}a=1+

7

To find,

The value of \dfrac{-6}{a}

a

−6

.

Solution,

To find the value of \dfrac{-6}{a}

a

−6

, we put the value of a as :

\dfrac{-6}{a}=\dfrac{-6}{1+\sqrt{7} }

a

−6

=

1+

7

−6

Rationalizing by multiplying and dividing by 1-\sqrt{7}1−

7

. So,

\begin{gathered}\dfrac{-6}{a}=\dfrac{-6}{1+\sqrt{7} }\times \dfrac{1-\sqrt7}{1-\sqrt7}\\\\=\dfrac{-6(1-\sqrt7)}{1^2-(\sqrt{7})^2 }\\\\=\dfrac{-6+6\sqrt7}{1-7}\\\\=\dfrac{-6+6\sqrt7}{-6}\\\\=\dfrac{-6(1-\sqrt7)}{-6}\\\\=1-\sqrt7\end{gathered}

a

−6

=

1+

7

−6

×

1−

7

1−

7

=

1

2

−(

7

)

2

−6(1−

7

)

=

1−7

−6+6

7

=

−6

−6+6

7

=

−6

−6(1−

7

)

=1−

7

So, the value of \dfrac{-6}{a}

a

−6

is equal to 1-\sqrt{7}1−

7