Question

If a = $\dfrac{3 + \sqrt{5}}{2}$ Then find,
$\sf \bigg \lgroup {a}^{2} + \dfrac{1}{ {a}^{2} } \bigg \rgroup$
​

Answer 1

Answer:

$The \: answer \: \: is \: \: 7$

Solution:

a = 3 +∫5 / 2

Therefore, 1 / a = 2 / 3 + ∫5

= 2 * ( 3 - √5 ) / ( 3 + ∫5 ) * ( 3 - ∫ 5 )

= 2 * ( 3 - √5 ) / ( 3 ) ^ 2 - ( √5 ) ^ 2

= 3 - √5 / 2

a + 1 /a = 3+√5 /2 + 3-√5 /2

= 6 / 2

= 3

( a + 1 /a ) ^ 2 = a^2 + 1/a^2 + 2 (a)(1/a)

= ( a + 1 /a )^2 + 2 (a)(1/a)

= ( 3) ^2 - 2 = ( a + 1 /a )^2 + 2 (a)(1/a) - 2 ( Adding and subtracting 2 on both sides )

9 -2 = ( a + 1 /a )^2

= 7 .

Therefore, (a + 1/a)^2 = 7

Answer 2

Assume we already know the value and say $a+\dfrac{1}{a} =t$.

The equation is $a^2-ta+1=0$, which will be replaced by $x$.

Assume, $x^2-tx+1=0$ has $a$ and $b$ as solutions.

Now here we use Vieta's formula;

• Product of roots: 1

The other solution is the multiplicative inverse.

$\therefore a=\dfrac{3+\sqrt{5}}{2} , b=\dfrac{3-\sqrt{5} }{2}$

Now finding sum and product,

$\begin{cases} & a+b=3 \\ & ab=1 \end{cases}$

Using Vieta's formula again, we constructed an equation.

$x^2-3x+1=0$

Dividing by $x$,

$x-3+\dfrac{1}{x} =0$ $\therefore x+\dfrac{1}{x} =3$

Now since we know $x=a$ is a solution, we can find it!

$(x+\dfrac{1}{x} )^2=3^2$ $\therefore x^2+\dfrac{1}{x^2} =7$

Conclusion

The required solution is

$a^2+\dfrac{1}{a^2} =7$

Learn More

Vieta's formula says the product and sum of the roots can be found using coefficients.

Say we have a quadratic equation $ax^2+bx+c=0$ that zeros are $\alpha ,\beta$.

$\therefore ax^2+bx+c=a(x-\alpha )(x-\beta )$

Now it is easy to find product and sum using coefficient comparison method.

Similarly, this method works to higher degrees.