Question

If x>1, then (1-x)/(1-sqrt. x)=
A. x
B. sqrt. x
C. x- sqrt. x
D. 1+sqrt. x
E 1- sqrt. x

Answer 1

\[(1-x)/(1-\sqrt{x})=\]

Answer 2

\[\large \frac{1-x}{1-\sqrt{x}}=\frac{1-x}{1-\sqrt{x}}\times \frac{1+\sqrt{x}}{1+\sqrt{x}}\]
\[\large =\frac{(1-x)(1+\sqrt{x})}{1-x}\]
\[\large =1+\sqrt{x}\]

Answer 3

why did you multiple it by 1+ sqrt. x/ 1+ sqrt. x ????

Answer 4

That's called rationalising the denominator.

Answer 5

You usually do that to get rid of square root signs. You multiply the denominator with it's conjugate.

Answer 6

but sign is different?

Answer 7

I already explained it to you in the second sentence?

Answer 8

nooo, i mean it is +, cause it is not 1+ sqrt. x, but 1- sqrt. x?

Answer 9

You multiply by its conjugate in this scenario because it only has one term under a square root sign. It's not all under a square root sign. For example:
\[\large \frac{1+x}{\sqrt{1+x}}\]
You can rationalise the denominator by multiplying the denominator with the same thing.
But for questions like the above, you use the property of "difference between two squares".
For example:
\[\large (x-1)(x+1)=x^2-1\]
And if it was square root x:
\[\large (\sqrt{x}-1)(\sqrt{x}+1)=(\sqrt{x})(\sqrt{x})-(1)(1)\]
\[\large =x-1\]

Answer 10

The conjugate of \[1+\sqrt{x}\] is \[1-\sqrt{x}\]

Answer 11

When you multiply by it's conjuagte, you get rid of the square root sign instantly.

Answer 12

OK, understand. thank you very much :)))