Question

How to rationalize the denominator of 5-3√14/7+2√14

Answer 1

GIVEN :

How to rationalize the denominator of $\frac{5-3\sqrt{14}}{7+2\sqrt{14}}$

TO FIND :

The rationalize the denominator of the given expression $\frac{5-3\sqrt{14}}{7+2\sqrt{14}}$

SOLUTION :

Given that the expression is $\frac{5-3\sqrt{14}}{7+2\sqrt{14}}$

Now rationalize the expression as below,

$\frac{5-3\sqrt{14}}{7+2\sqrt{14}}$

Multiply and dividing by the denominator's conjugate we get,

$=\frac{5-3\sqrt{14}\times (7-2\sqrt{14})}{(7+2\sqrt{14})\times (7-2\sqrt{14})}$

By using the Distributive property :

(a+b)x=ax+ay

By using the Algebraic identity :

$(a+b)(a-b)=a^2-b^2$

$=\frac{5(7)+5(-2\sqrt{14})+(-3\sqrt{14})(7)+(-3\sqrt{14})(-2\sqrt{14})}{7^2-(2\sqrt{14})^2}$

$=\frac{35-10\sqrt{14}-21\sqrt{14}+6(14)}{49-(-2^2)(\sqrt{14})^2}$

By using the property of square root :

$\sqrt{a}^2=a$

$=\frac{35-31\sqrt{14}+84}{49-(4)(14)}$

By adding the like terms,

$=\frac{119-31\sqrt{14}}{49-56}$

$=\frac{119-31\sqrt{14}}{-7}$

$=\frac{-119+31\sqrt{14}}{7}$

or $=\frac{31\sqrt{14}-119}{7}$

∴ $\frac{5-3\sqrt{14}}{7+2\sqrt{14}}=\frac{-119+31\sqrt{14}}{7}$

or $\frac{5-3\sqrt{14}}{7+2\sqrt{14}}=\frac{31\sqrt{14}-119}{7}$

∴ the given expression the rationalized denominator becomes  $\frac{-119+31\sqrt{14}}{7}$ or $\frac{31\sqrt{14}-119}{7}$.

Answer 2

Step-by-step explanation:

GIVEN :

How to rationalize the denominator of \frac{5-3\sqrt{14}}{7+2\sqrt{14}}

7+2

14

5−3

14

TO FIND :

The rationalize the denominator of the given expression \frac{5-3\sqrt{14}}{7+2\sqrt{14}}

7+2

14

5−3

14

SOLUTION :

Given that the expression is \frac{5-3\sqrt{14}}{7+2\sqrt{14}}

7+2

14

5−3

14

Now rationalize the expression as below,

\frac{5-3\sqrt{14}}{7+2\sqrt{14}}

7+2

14

5−3

14

Multiply and dividing by the denominator's conjugate we get,

=\frac{5-3\sqrt{14}\times (7-2\sqrt{14})}{(7+2\sqrt{14})\times (7-2\sqrt{14})}=

(7+2

14

)×(7−2

14

)

5−3

14

×(7−2

14

)

By using the Distributive property :

(a+b)x=ax+ay

By using the Algebraic identity :

(a+b)(a-b)=a^2-b^2(a+b)(a−b)=a

2

−b

2

=\frac{5(7)+5(-2\sqrt{14})+(-3\sqrt{14})(7)+(-3\sqrt{14})(-2\sqrt{14})}{7^2-(2\sqrt{14})^2}=

7

2

−(2

14

)

2

5(7)+5(−2

14

)+(−3

14

)(7)+(−3

14

)(−2

14

)

=\frac{35-10\sqrt{14}-21\sqrt{14}+6(14)}{49-(-2^2)(\sqrt{14})^2}=

49−(−2

2

)(

14

)

2

35−10

14

−21

14

+6(14)

By using the property of square root :

\sqrt{a}^2=a

a

2

=a

=\frac{35-31\sqrt{14}+84}{49-(4)(14)}=

49−(4)(14)

35−31

14

+84

By adding the like terms,

=\frac{119-31\sqrt{14}}{49-56}=

49−56

119−31

14

=\frac{119-31\sqrt{14}}{-7}=

−7

119−31

14

=\frac{-119+31\sqrt{14}}{7}=

7

−119+31

14

or =\frac{31\sqrt{14}-119}{7}=

7

31

14

−119

∴ \frac{5-3\sqrt{14}}{7+2\sqrt{14}}=\frac{-119+31\sqrt{14}}{7}

7+2

14

5−3

14

=

7

−119+31

14

or \frac{5-3\sqrt{14}}{7+2\sqrt{14}}=\frac{31\sqrt{14}-119}{7}

7+2

14

5−3

14

=

7

31

14

−119

∴ the given expression the rationalized denominator becomes \frac{-119+31\sqrt{14}}{7}

7

−119+31

14

or \frac{31\sqrt{14}-119}{7}

7

31

14

−119

.