If two irrational number are given as a = $\dfrac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }}$ and b = $\dfrac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }}$, If value of a + b -5ab is -4$\sqrt {\text{m}} $+ 5 then find m.
Answer
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Hint:- Take LCM while solving the equation and simplify before equating. Equate irrational terms of both sides.
Complete step-by-step solution -
Given, a = $\dfrac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }}$ and b = $\dfrac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }}$ and,
a + b -5ab = -4$\sqrt {\text{m}} $+ 5
Putting the values of a and b in above equation
$\dfrac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }}$ + $\dfrac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }}$ - 5 $\left( {\dfrac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }}} \right)$$\left( {\dfrac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }}} \right)$ = -4$\sqrt {\text{m}} $+ 5
Taking L.C.M. , we get
$\dfrac{{{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2} + {{\left( {\sqrt 3 + \sqrt 2 } \right)}^2} - 5\left( {\sqrt 3 - \sqrt 2 } \right)\left( {\sqrt 3 + \sqrt 2 } \right)}}{{\left( {\sqrt 3 - \sqrt 2 } \right)\left( {\sqrt 3 + \sqrt 2 } \right)}}$= -4$\sqrt {\text{m}} $+ 5
Using the algebraic identities ${{\text{(a + b)}}^2}{\text{ = }}{{\text{a}}^2}{\text{ + }}{{\text{b}}^2}{\text{ + 2ab }}$and ${{\text{(a - b)}}^2}{\text{ = }}{{\text{a}}^2}{\text{ + }}{{\text{b}}^2}{\text{ - 2ab }}$, we get
$\dfrac{{\left( {3 + 2 - 2\sqrt 6 } \right) + \left( {3 + 2 + 2\sqrt 6 } \right) - 5\left( {\sqrt 3 - \sqrt 2 } \right)\left( {\sqrt 3 + \sqrt 2 } \right)}}{{\left( {\sqrt 3 - \sqrt 2 } \right)\left( {\sqrt 3 + \sqrt 2 } \right)}}$= -4$\sqrt {\text{m}} $+ 5
Simplifying and using another algebraic identity $\left( {{\text{a + b}}} \right)\left( {{\text{a - b}}} \right) = {{\text{a}}^2}{\text{ - }}{{\text{b}}^2}$
$\dfrac{{10 - 5\left( {3 - 2} \right)}}{{\left( {3 - 2} \right)}}$= -4$\sqrt {\text{m}} $+ 5
$5$= -4$\sqrt {\text{m}} $+ 5
-4$\sqrt {\text{m}} $= 0
$\sqrt {\text{m}} $=0
m = 0.
Hence, m = 0.
Note:- In these types of questions, we need to simplify and apply all possible algebraic identity. The identity $\left( {{\text{a + b}}} \right)\left( {{\text{a - b}}} \right) = {{\text{a}}^2}{\text{ - }}{{\text{b}}^2}$ is very helpful in simplification.
Complete step-by-step solution -
Given, a = $\dfrac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }}$ and b = $\dfrac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }}$ and,
a + b -5ab = -4$\sqrt {\text{m}} $+ 5
Putting the values of a and b in above equation
$\dfrac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }}$ + $\dfrac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }}$ - 5 $\left( {\dfrac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }}} \right)$$\left( {\dfrac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }}} \right)$ = -4$\sqrt {\text{m}} $+ 5
Taking L.C.M. , we get
$\dfrac{{{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2} + {{\left( {\sqrt 3 + \sqrt 2 } \right)}^2} - 5\left( {\sqrt 3 - \sqrt 2 } \right)\left( {\sqrt 3 + \sqrt 2 } \right)}}{{\left( {\sqrt 3 - \sqrt 2 } \right)\left( {\sqrt 3 + \sqrt 2 } \right)}}$= -4$\sqrt {\text{m}} $+ 5
Using the algebraic identities ${{\text{(a + b)}}^2}{\text{ = }}{{\text{a}}^2}{\text{ + }}{{\text{b}}^2}{\text{ + 2ab }}$and ${{\text{(a - b)}}^2}{\text{ = }}{{\text{a}}^2}{\text{ + }}{{\text{b}}^2}{\text{ - 2ab }}$, we get
$\dfrac{{\left( {3 + 2 - 2\sqrt 6 } \right) + \left( {3 + 2 + 2\sqrt 6 } \right) - 5\left( {\sqrt 3 - \sqrt 2 } \right)\left( {\sqrt 3 + \sqrt 2 } \right)}}{{\left( {\sqrt 3 - \sqrt 2 } \right)\left( {\sqrt 3 + \sqrt 2 } \right)}}$= -4$\sqrt {\text{m}} $+ 5
Simplifying and using another algebraic identity $\left( {{\text{a + b}}} \right)\left( {{\text{a - b}}} \right) = {{\text{a}}^2}{\text{ - }}{{\text{b}}^2}$
$\dfrac{{10 - 5\left( {3 - 2} \right)}}{{\left( {3 - 2} \right)}}$= -4$\sqrt {\text{m}} $+ 5
$5$= -4$\sqrt {\text{m}} $+ 5
-4$\sqrt {\text{m}} $= 0
$\sqrt {\text{m}} $=0
m = 0.
Hence, m = 0.
Note:- In these types of questions, we need to simplify and apply all possible algebraic identity. The identity $\left( {{\text{a + b}}} \right)\left( {{\text{a - b}}} \right) = {{\text{a}}^2}{\text{ - }}{{\text{b}}^2}$ is very helpful in simplification.
Last updated date: 17th Sep 2023
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