Question

Rationalize the denominator

7/√7-√3

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Answer 1

Answer: (7√7 + 7√3)/4

Explanation:

The conjugate of (√7 - √3) is (√7 + √3).

Step 1: Multiply the numerator and denominator by the conjugate (√7 + √3):

7 * (√7 + √3) / (√7 - √3) * (√7 + √3)

Step 2: Apply the distributive property to simplify the numerator:

7√7 + 7√3 / (√7)^2 - (√3)^2

Step 3: Simplify the denominator:

7√7 + 7√3 / 7 - 3

Step 4: Simplify further if possible:

7√7 + 7√3 / 4

So, the rationalized form of 7/(√7 - √3) is (7√7 + 7√3)/4.

Answer 2

To rationalize the denominator of the expression $\frac{7}{\sqrt{7}-\sqrt{3}}$, we can use the conjugate of the denominator. The conjugate of $\sqrt{7}-\sqrt{3}$ is $\sqrt{7}+\sqrt{3}$.

We multiply the numerator and denominator by the conjugate:

$$\frac{7}{\sqrt{7}-\sqrt{3}} \times \frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}+\sqrt{3}}$$

Expanding the numerator and denominator:

$$\frac{7(\sqrt{7}+\sqrt{3})}{(\sqrt{7}-\sqrt{3})(\sqrt{7}+\sqrt{3})}$$

Simplifying the denominator:

$$\frac{7(\sqrt{7}+\sqrt{3})}{(\sqrt{7})^2 - (\sqrt{3})^2}$$

This simplifies further:

$$\frac{7(\sqrt{7}+\sqrt{3})}{7-3}$$

And finally, we simplify the expression:

$$\frac{7(\sqrt{7}+\sqrt{3})}{4}$$

So, the rationalized form of $\frac{7}{\sqrt{7}-\sqrt{3}}$ is $\frac{7(\sqrt{7}+\sqrt{3})}{4}$.