Question

underoot 15 + underoot 17 ) underoot 15 - underoot 17​

Answer 1

Given to find the value of :-

$( \sqrt{15} + \sqrt{17} )( \sqrt{15} - \sqrt{17} )$

Solution:-

Here it is in form of ,

$\bigstar \underline{\boxed{ \mathfrak{(a+b)(a-b)=a^2-b^2}}}$

By using this identity we can solve.

$= ( \sqrt{15} ) { }^{2} - ( \sqrt{17} ) {}^{2}$

$= 15 - 17$

$= \: - 2$

So, the value of

$( \sqrt{15} + \sqrt{17} )( \sqrt{15} - \sqrt{17} ) = - 2$

Alternate method :-

$( \sqrt{15} + \sqrt{17} )( \sqrt{15} - \sqrt{17} )$

We can do normal multiplication

$\sqrt{15} ( \sqrt{15} - \sqrt{17} ) + \sqrt{17} ( \sqrt{15} - \sqrt{17} )$

$( \sqrt{15} ) {}^{2} - \sqrt{15} \times \sqrt{17} + \sqrt{15} \times \sqrt{17} - ( \sqrt{17} ) {}^{2}$

$( \sqrt{15} ) {}^{2} - ( \sqrt{17} ) {}^{2}$

$15 - 17$

$- 2$

Hence , in both methods same answer

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# Know more Algebraic Identities:-

(a+ b)² = a² + b² + 2ab

( a - b )² = a² + b² - 2ab

( a + b )² + ( a - b)² = 2a² + 2b²

( a + b )² - ( a - b)² = 4ab

( a + b + c )² = a² + b² + c² + 2ab + 2bc + 2ca

a² + b² = ( a + b)² - 2ab

(a + b )³ = a³ + b³ + 3ab ( a + b)

( a - b)³ = a³ - b³ - 3ab ( a - b)

If a + b + c = 0 then a³ + b³ + c³ = 3abc