Question

please answer b,d questions carefully plss and answer in front ​

Answer 1

Required to Know

• Rationalization of the Denominator

The rationalization of the denominator is literally converting the denominator to a rational number.

• Algebraic Identity

An identity is an equation that is always true regardless of the values of the unknowns.

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Solution

The question requires identity. Because in all the questions, the value we find is actually rational numbers, so we will use algebraic identity.

Q27.

After rationalization, the value of $\dfrac{1}{x} =3-\sqrt{8}$, so the answer is $\boxed{x+\dfrac{1}{x} =6}$.

Q28.

After rationalization, the value of $\dfrac{1}{x} =3+\sqrt{8}$, so the first required value is

$x-\dfrac{1}{x} =-2\sqrt{8}$.

After squaring both sides, we find that,

$(x+\dfrac{1}{x} )^2=(4\sqrt{2} )^2\implies (x+\dfrac{1}{x} )^2=32$

So the required answer is $\boxed{(x-\dfrac{1}{x} )^2=32}$.

Q29.

After rationalization, the value of $\dfrac{1}{x} =2-\sqrt{3}$, so the first required value is $x+\dfrac{1}{x} =4$.

After squaring both sides, we find that,

$(x+\dfrac{1}{x} )^2=4^2\implies x^2+2+\dfrac{1}{x^2} =16$

So the required answer is $\boxed{x^2+\dfrac{1}{x^2} =14}$.

Q30.

After rationalization, the value of $b=\dfrac{1}{a} =2+\sqrt{3}$, so the first required value is $a+b=4$.

After squaring both sides, we find that,

$\implies (a+b)^2=4^2$

$\implies a^2+2ab+b^2=16$

Since it is given that $2ab=\dfrac{2a}{a} =2$,

$\implies a^2+b^2=14$

So the required answer is $\boxed{a^2+b^2=14}$.

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Hey! Sorry for the late answer. Hope this helps and please read the text below!

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Mental Trick

How do you rationalize in 1 sec?

If we find that the square of each term in the denominator differs by 1, we can ignore the rationalization step.

For example, all the questions here have squares that differ by 1. If we apply identity here, the denominator is 1. Let's apply the trick.

$x=\dfrac{1}{12-\sqrt{143}}$

How do you rationalize it?

$x=\dfrac{1}{12-\sqrt{143}}\times \dfrac{12+\sqrt{143} }{12+\sqrt{143} } =\dfrac{12+\sqrt{143} }{144-143}=\boxed{12+\sqrt{143} }$

But, as we see that the squares differ by 1, we can use this trick.

$\boxed{x=12+\sqrt{143} }$