Question

if a² -3a -1= 0 find a²+1/a² solve the linear expansions​

Answer 1

Given:-

• a² - 3a - 1 = 0

To Find:-

• $\rm{a^{2} + 1/a^{2}}$

Now,

→ $\sf{ a^{2} - 3a - 1 = 0}$

→ $\sf{ a^{2} - 3a = 1}$

• Dividing it by a

→ $\sf{ a - 3 = \dfrac{1}{a}}$

→ $\sf{ -3 = \dfrac{1}{a} - a }$

• Squaring Both sides

→ $\sf{ (-3)^2 = (\dfrac{1}{a} - a )^2}$

→ $\sf{ 9 = a^2 +\dfrac{1} {a^2} - 2\times{a}\times{\dfrac{1} {a}}$

→ $\sf{ 9 = a^2 +\dfrac{1}{a^2} - 2\times{\cancel{a}}\times{\dfrac{1}{\cancel{a}}}$

→ $\sf{ 9 + 2 = a^2 +\dfrac{1}{a^2}}$

→ $\sf{ 11 = a^2 +\dfrac{1}{a^2}}$

Hence, The Value of a² + 1/a² is 11.

Answer 2

Answer:

GIVEN:

${a}^{2} - 3a - 1 = 0$

TO FIND:

a^2+1/a^2

SOLUTION:

a^2-3a-1=0

=>a^2-3a=1

DIVIDING BOTH SIDES BY a.

1/a(a^2-3a)=1/a

=>a-3=1/a

=>a=1/a+3

=>-3=1/a-a

SQUARING BOTH SIDES:

$({ - 3})^{2} = ({ \frac{1}{a} - a) }^{2}$

=9+2=1/a^2+a^2

=11=1/a^2+a^2 (ANSWER)