Answer :

sitapoudellamichhane

Answer:

To find \(a\) and \(b\) such that \(\frac{4 + 2\sqrt{5}}{4 - 2\sqrt{5}} = a + b\sqrt{5}\):

1. Multiply the numerator and the denominator by the conjugate of the denominator, \(4 + 2\sqrt{5}\):

\[

\frac{(4 + 2\sqrt{5})(4 + 2\sqrt{5})}{(4 - 2\sqrt{5})(4 + 2\sqrt{5})}

\]

2. Simplify the denominator:

\[

(4 - 2\sqrt{5})(4 + 2\sqrt{5}) = 4^2 - (2\sqrt{5})^2 = 16 - 20 = -4

\]

3. Simplify the numerator:

\[

(4 + 2\sqrt{5})^2 = 16 + 16\sqrt{5} + 20 = 36 + 16\sqrt{5}

\]

4. Divide by \(-4\):

\[

\frac{36 + 16\sqrt{5}}{-4} = -9 - 4\sqrt{5}

\]

Thus, \(a = -9\) and \(b = -4\).

semsee45

Answer:

[tex]-9-4\sqrt{5}[/tex]

Step-by-step explanation:

Given rational expression:

[tex]\dfrac{4+2\sqrt{5}}{4-2\sqrt{5}}[/tex]

To express the given expression in the form [tex]a+b\sqrt{5}[/tex], we need to rationalize the denominator.

To do this, multiply the numerator and denominator by the conjugate of the denominator, which is [tex]4+2\sqrt{5}[/tex]:

[tex]\dfrac{\left(4+2\sqrt{5}\right)\left(4+2\sqrt{5}\right)}{\left(4-2\sqrt{5}\right)\left(4+2\sqrt{5}\right)}[/tex]

Simplify by expanding the brackets of the numerator and denominator:

[tex]\dfrac{4 \cdot 4 +4 \cdot 2\sqrt{5}+4 \cdot 2\sqrt{5}+2\sqrt{5} \cdot 2\sqrt{5}}{4 \cdot 4 +4 \cdot 2\sqrt{5}-4 \cdot 2\sqrt{5}-2\sqrt{5} \cdot 2\sqrt{5}}[/tex]

[tex]\dfrac{16 +8\sqrt{5}+8\sqrt{5}+4\sqrt{5} \sqrt{5}}{16 +8\sqrt{5}-8\sqrt{5}-4\sqrt{5}\sqrt{5}}\\\\\\\\\dfrac{16 +8\sqrt{5}+8\sqrt{5}+4\cdot 5}{16 +8\sqrt{5}-8\sqrt{5}-4\cdot 5}\\\\\\\\\dfrac{16 +8\sqrt{5}+8\sqrt{5}+20}{16 +8\sqrt{5}-8\sqrt{5}-20}\\\\\\\\\dfrac{16 +16\sqrt{5}+20}{16 -20}\\\\\\\\\dfrac{36 +16\sqrt{5}}{-4}\\\\\\\\\dfrac{36}{-4}+\dfrac{16\sqrt{5}}{-4}\\\\\\\\-9-4\sqrt{5}[/tex]

Therefore, the given expression simplifies to:

[tex]\Large\boxed{\boxed{-9-4\sqrt{5}}}[/tex]

This means that:

  • a = -9
  • b = -4

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