Question

if a square minus 5 + 1 equal to zero and a not equal to zero find a square + 1 by a square​

Answer 1

The value is $a^2+\frac{1}{a^2}=23.08$.

Step-by-step explanation:

Given : Equation $a^2-5a+1=0$ and a not equal to zero.

To find : The value of  $a^2+\frac{1}{a^2}$ ?

Solution :

Quadratic equation, $a^2-5a+1=0$

Solve by quadratic formula, $x=\frac{-B\pm\sqrt{B^2-4AC}}{2A}$

Here, A=1, B=-5 and C=1

$a=\frac{-(-5)\pm\sqrt{(-5)^2-4(1)(1)}}{2(1)}$

$a=\frac{5\pm\sqrt{21}}{2}$

$a=\frac{5+\sqrt{21}}{2},\frac{5-\sqrt{21}}{2}$

$a=4.79,0.20$

Reject a=0.20 as it approx to 0.

So, the value of a≈4.8

Substitute in the value,

$a^2+\frac{1}{a^2}=(4.8)^2+\frac{1}{(4.8)^2}$

$a^2+\frac{1}{a^2}=23.04+\frac{1}{23.04}$

$a^2+\frac{1}{a^2}=23.04+\frac{1}{23.04}$

$a^2+\frac{1}{a^2}=23.04+0.04$

$a^2+\frac{1}{a^2}=23.08$

Therefore, the value is $a^2+\frac{1}{a^2}=23.08$.

#Learn more

3x^2-2x-1 quadratic equation ​

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