Question

if x=7+ root 48,find the value of root x +1/root x ​

Answer 1

Answer:

Step-by-step explanation:

x=7+√40

x=7+2√10

 =(√5)²+(√2)²+2×√2×√5

 =(√5+√2)²

√x=√5+√2

1/√x=1/√5+√2

     =√5-√2/3

√x+1/√x=√5+√2 + √5-√2/3

           =4√5+2√2/3

           ie 2/3( 2√5+√2)

hope this is helpful

Answer 2

The value is $\sqrt{x}+\frac{1}{\sqrt{x}}=4$.

Step-by-step explanation:

Given : If $x=7+\sqrt{48}$

To find : The value of $\sqrt{x}+\frac{1}{\sqrt{x}}$ ?

Solution :

$x=7+\sqrt{48}$

Write it as,

$x=7+4\sqrt{3}$

$x=(2)^2+2\times 2\times \sqrt{3}+(\sqrt{3})^2$

$x=(2+\sqrt{3})^2$

Now taking root both side,

$\sqrt{x}=\sqrt{(2+\sqrt{3})^2}$

$\sqrt{x}=2+\sqrt{3}$ .....(1)

Now,

$\frac{1}{\sqrt{x}}=\frac{1}{2+\sqrt3}$

Rationalize the denominator,

$\frac{1}{\sqrt{x}}=\frac{1}{2+\sqrt3}\times \frac{2-\sqrt3}{2-\sqrt3}$

$\frac{1}{\sqrt{x}}=\frac{2-\sqrt3}{2^2-(\sqrt3)^2}$

$\frac{1}{\sqrt{x}}=\frac{2-\sqrt3}{4-3}$

$\frac{1}{\sqrt{x}}=2-\sqrt3$   ....(2)

Add (1) and (2),

$\sqrt{x}+\frac{1}{\sqrt{x}}=2+\sqrt3+2-\sqrt3$

$\sqrt{x}+\frac{1}{\sqrt{x}}=4$

Therefore, the value is $\sqrt{x}+\frac{1}{\sqrt{x}}=4$

#Learn more

If x=7+root 40 find the value of root x +1/root x

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