Question

if a2-3a+1=0 so, a4+1÷a4=?

Answer 1

a(2-3) = -1
a(-1) = -1
so, a = 1
Then, (1)4+1/(1)4 = 17/4

Answer 2

Correct Question :-

If a² - 3a + 1 = 0, find the value of $\sf a^4 + \dfrac{1}{a^4}$

Answer :-

$\sf a^4 + \dfrac{1}{{a}^{4}} = 47$

Solution :-

First find the value of a + 1/a

a² - 3a + 1 = 0

⇒ a² - 3a = - 1

⇒ a(a - 3) = -1

⇒ a - 3 = - 1/a

⇒ a - 3 + 1/a = 0

⇒ a + 1/a = 3

Squaring on both sides

(a + 1/a)² = (3)²

⇒ (a + 1/a)² = 9

⇒ (a)² + (1/a)² + 2(a)(1/a) = 9

[Since (a + b)² = a² + b² + 2ab]

⇒ a² + 1²/a² + 2 = 9

⇒ a² + 1/a² + 2 = 9

⇒ a² + 1/a² = 9 - 2

⇒ a² + 1/a² = 7

Squaring on both sides

(a² + 1/a²)² = (7)²

⇒ (a² + 1/a²)² = 49

⇒ (a²)² + (1/a²)² + 2(a²)(1/a²) = 49

[Since (a + b)² = a² + b² + 2ab]

$\sf \implies a^{2(2)} + \dfrac{ {1}^{2} }{ {( {a}^{2})}^{2} } + 2 = 49$

$\sf \implies a^4 + \dfrac{1}{{a}^{2(2)}} + 2 = 49$

$\sf \implies a^4 + \dfrac{1}{{a}^{4}} + 2 = 49$

$\sf \implies a^4 + \dfrac{1}{{a}^{4}} = 49 - 2$

$\sf \implies a^4 + \dfrac{1}{{a}^{4}} = 47$

$\bf \therefore a^4 + \dfrac{1}{{a}^{4}} = 47$

Identity used :-

• (a + b)² = a² + b² + 2ab