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655-705 Level|   Algebra|   Roots|      
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Bunuel
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What is the value of the expression below: 11 Jan 2022, 03:45
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65% (02:48) correct 35% (02:45) wrong based on 273 sessions

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What is the value of the expression below:

\(\frac{7\sqrt{3}}{\sqrt{10} + \sqrt{3}}-\frac{2\sqrt{5}}{\sqrt{6} + \sqrt{5}}-\frac{3\sqrt{2}}{\sqrt{15} + 3\sqrt{2}}\)

A. -1
B. 1
C. 6
D. 12
E. 54


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B
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What is the value of the expression below: 17 Jan 2022, 12:46
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Core concept is to multiply the denominator by the numerator and to realize (a+b) * (a-b) = a^2 + b^2. This will give you a whole number usually in the denominator for these questions making it easy to solve!

Hope it helps.
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What is the value of the expression below: 17 Jan 2022, 22:34
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What is the value of the expression below:

\(\frac{7\sqrt{3}}{\sqrt{10} + \sqrt{3}}-\frac{2\sqrt{5}}{\sqrt{6} + \sqrt{5}}-\frac{3\sqrt{2}}{\sqrt{15} + 3\sqrt{2}}\)

A. -1
B. 1
C. 6
D. 12
E. 54


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The ideal way is to rationalise each term and then simplify. For example:

We can rationalise the term
7√3/(√10 + √3)
by multiplying numerator and denominator by (√10 - √3)
That would remove the root-terms from the denominator.
Similarly we can work with each of the other 2 terms.

However, this would be a lengthy process.

The alternative is to use approximations intelligently.

For example, √10 ~ √9 = 3
The actual value of √10 is 3.16. So, the error isn't too much. Also, √10 is in the denominator, so the effect is reduced even more.

Thus, approximating the terms, we can proceed as I have shown in the attachment below:

Attachment:
IMG_20220118_110250__01.jpg
IMG_20220118_110250__01.jpg [ 1.7 MiB | Viewed 4621 times ]

Answer B

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What is the value of the expression below: 26 Jul 2025, 05:20
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We need to simplify \(\frac{7\sqrt{3}}{\sqrt{10} + \sqrt{3}}-\frac{2\sqrt{5}}{\sqrt{6} + \sqrt{5}}-\frac{3\sqrt{2}}{\sqrt{15} + 3\sqrt{2}}\)

We will be using a concept called as Rationalizing Roots, we will multiply the numerator and the denominator with the conjugate of the denominator
Conjugate of \( \sqrt{10} + \sqrt{3} \) = \( \sqrt{10} - \sqrt{3} \)
Conjugate of \( \sqrt{6} + \sqrt{5} \) = \( \sqrt{6} - \sqrt{5} \)
Conjugate of \( \sqrt{15} + 3\sqrt{2} \) = \( \sqrt{15} - 3\sqrt{2} \)

=> \( \frac{7\sqrt{3}}{\sqrt{10} + \sqrt{3}} * \frac{\sqrt{10} - \sqrt{3}}{\sqrt{10} - \sqrt{3}} - \frac{2\sqrt{5}}{\sqrt{6} + \sqrt{5}} * \frac{\sqrt{6} - \sqrt{5}}{ \sqrt{6} - \sqrt{5}} - \frac{3\sqrt{2}}{\sqrt{15} + 3\sqrt{2}} * \frac{\sqrt{15} - 3\sqrt{2}}{\sqrt{15} - 3\sqrt{2}} \)

Now, in each of the terms in the denominator we have (a+b) * (a-b) = \(a^2 - b^2\)

=> \(\frac{7\sqrt{3} * (\sqrt{10} - \sqrt{3})}{(\sqrt{10})^2 - (\sqrt{3})^2} - \frac{2\sqrt{5} * ( \sqrt{6} - \sqrt{5})}{(\sqrt{6})^2 - (\sqrt{5})^2} - \frac{3\sqrt{2} * ( \sqrt{15} - 3\sqrt{2} )}{(\sqrt{15})^2 - (3\sqrt{2})^2}\)
=> \( \frac{7 (\sqrt{3*10} - \sqrt{3^2})}{10 - 3} - \frac{(2\sqrt{5*6} - 2 \sqrt{5*5})}{6 - 5} - \frac{3(\sqrt{2*15} - 3\sqrt{2*2})}{15 - 9*2}\)
=> \( \frac{7 (\sqrt{30} - 3)}{7} - \frac{(2\sqrt{30} - 2*5)}{1} - \frac{3(\sqrt{30} - 3*2)}{15 - 18}\)
=> \((\sqrt{30} - 3) - (2\sqrt{30} - 10 ) - \frac{3(\sqrt{30} - 6)}{-3}\)
=> \(\sqrt{30} - 3 - 2\sqrt{30} + 10 + \sqrt{30} - 6\)
=> 10 - 9 = 1

So, Answer will be B
Hope it helps!

Playlist on Solved Problems on Roots here

Watch the following video to MASTER Rationalizing Roots


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