Question

if a square minus 3a + 1 equals to zero find the value of a square + 1 by a square​

Answer 1

Given : a²-3a+1 =0

To find : $a^2+\frac{1}{a^2}$

Solution:

a²-3a+1 =0

a²-3a=-1

a(a-3) = -1

$a-3 = \frac{-1}{a}\\\\a-3+\frac{1}{a}=0\\\\a+\frac{1}{a}=3..(1)$

now ,

squaring both sides on equation 1

we get

$(a+\frac{1}{a})^2=3^2\\\\a^2+\frac{1}{a^2}+2\times a \times \frac{1}{a}=9\\\\a^2+\frac{1}{a^2}+2=9\\\\a^2+\frac{1}{a^2}= 9-2 \\\\a^2+\frac{1}{a^2} = 7$

hence , The value if $a^2+\frac{1}{a^2}$  is 7

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Answer 2

Answer:

hope it is helpful

Step-by-step explanation:

Given : a²-3a+1 =0

To find : a^2+\frac{1}{a^2}a

2

+

a

2

1

Solution:

a²-3a+1 =0

a²-3a=-1

a(a-3) = -1

\begin{gathered}a-3 = \frac{-1}{a}\\\\a-3+\frac{1}{a}=0\\\\a+\frac{1}{a}=3..(1)\end{gathered}

a−3=

a

−1

a−3+

a

1

=0

a+

a

1

=3..(1)

now ,

squaring both sides on equation 1

we get

\begin{gathered}(a+\frac{1}{a})^2=3^2\\\\a^2+\frac{1}{a^2}+2\times a \times \frac{1}{a}=9\\\\a^2+\frac{1}{a^2}+2=9\\\\a^2+\frac{1}{a^2}= 9-2 \\\\a^2+\frac{1}{a^2} = 7\end{gathered}

(a+

a

1

)

2

=3

2

a

2

+

a

2

1

+2×a×

a

1

=9

a

2

+

a

2

1

+2=9

a

2

+

a

2

1

=9−2

a

2

+

a

2

1

=7

hence , The value if a^2+\frac{1}{a^2}a

2

+

a

2

1

is 7