Question

How do you simplify (3+square root of 2)(3-square root of 2) ?

Answer 1

Hint: The value of the algebraic expression $ \left( a+b \right)\left( a-b \right) $ is equal to $ {{a}^{2}}-{{b}^{2}} $ . The above question is in the format of $ \left( a+b \right)\left( a-b \right) $ we can solve the question by this expression by taking a equal to 3 and b equal to $ \sqrt{2} $

Complete step by step answer:
In the given question we have to find the value of (3+square root of 2)(3-square root of 2)
 $ \left( 3+\sqrt{2} \right)\left( 3-\sqrt{2} \right) $
We know that $ \left( a+b \right)\left( a-b \right) $ is equal to $ {{a}^{2}}-{{b}^{2}} $ .If we assume a is equal to 3 and b is equal to $ \sqrt{2} $ the answer to $ \left( 3+\sqrt{2} \right)\left( 3-\sqrt{2} \right) $ will be
 $ \left( {{3}^{2}}-{{\left( \sqrt{2} \right)}^{2}} \right) $ …….eq1
We know the property $ {{a}^{m}}{{b}^{m}} $ = $ {{\left( ab \right)}^{m}} $
We can write square root of 2 as $ {{2}^{\dfrac{1}{2}}} $
Applying above property to calculate $ \sqrt{2}\times \sqrt{2} $
So $ {{2}^{\dfrac{1}{2}}}\times {{2}^{\dfrac{1}{2}}} $ will be equal to $ {{\left( 2\times 2 \right)}^{\dfrac{1}{2}}} $
 $ \Rightarrow $ $ {{\left( 2\times 2 \right)}^{\dfrac{1}{2}}} $ = $ {{4}^{\dfrac{1}{2}}} $
We know that $ {{4}^{\dfrac{1}{2}}} $ is equal to 2.
So the square of $ \sqrt{2} $ is 2.
We can replace square of $ \sqrt{2} $ with 2 in eq1
 $ \Rightarrow 9-2=7 $
We can solve this by algebraic multiplication.
We can write $ \left( a+b \right)c $ as $ ac+bc $ .We will apply this property to solve the equation.
(3+square root of 2)(3-square root of 2)
= $ \left( 3+\sqrt{2} \right)\left( 3-\sqrt{2} \right) $
 $ \Rightarrow 3\left( 3-\sqrt{2} \right)+\sqrt{2}\left( 3-\sqrt{2} \right) $
 $ \Rightarrow 9+3\sqrt{2}-3\sqrt{2}-2 $
 $ =7 $

Note:
It is always good to remember some standard algebraic formula like $ \left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}} $ so that this type of problem would be easier to solve. Sometimes the problem will not be in a standard algebraic form in that case we always can use manual multiplication like we did in the second method. Remember the exponential property some of the method that might come into use are $ {{a}^{m}}\times {{a}^{n}}={{a}^{m+n}} $ , $ \dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}} $ , $ {{a}^{m}}{{b}^{m}}={{\left( ab \right)}^{m}} $ , and $ \dfrac{{{a}^{m}}}{{{b}^{m}}}={{\left( \dfrac{a}{b} \right)}^{m}} $ .