Mathematics: Question 1 Rationalise the denominator. 9 √7-2

the first answer would be Cthe next 3 questions are ACB...

Question 1 Rationalise the denominator. 9 √7-2

Ответ:

sarai47

16.03.2022

Mathematics

$\displaystyle \frac{9}{\sqrt{7}-2}=3\sqrt{7}+6$

Step-by-step explanation:

Rationalizing the Denominator

Rationalizing the denominator means to get rid of any radicals in the denominator.

We use the conjugate of the denominator to rationalize. The conjugate of a binomial is obtained by changing the sign that is between the two terms while keeping the same order of the terms. For example, the conjugate of

$\sqrt{7}-2$ is $\sqrt{7}+2$ .

We need to use the algebraic identity:

(a-b)(a+b)=a^2-b^2

Let's rationalize

$\displaystyle \frac{9}{\sqrt{7}-2}$

Multiply numerator and denominator by the conjugate of the denominator:

$\displaystyle \frac{9}{\sqrt{7}-2}=\frac{9}{\sqrt{7}-2}\cdot\frac{\sqrt{7}+2}{\sqrt{7}+2}$

Operate in the numerator and apply the identity in the denominator

$\displaystyle \frac{9}{\sqrt{7}-2}=\frac{9(\sqrt{7}+2)}{(\sqrt{7}-2)(\sqrt{7}+2)}$

$\displaystyle \frac{9}{\sqrt{7}-2}=\frac{9\sqrt{7}+18}{(\sqrt{7})^2-2^2}$

$\displaystyle \frac{9}{\sqrt{7}-2}=\frac{9\sqrt{7}+18}{7-4}=\frac{9\sqrt{7}+18}{3}$

Dividing:

$\boxed{\displaystyle \frac{9}{\sqrt{7}-2}=3\sqrt{7}+6}$

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Question 1
Rationalise the denominator.
9
√7-2

Ответ:

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