Question

X^4 - 142 X^2 + 1 =0 then the value of x?

Answer 1

Answer:

The values of x are $x=\sqrt{71+12\sqrt{35}},-\sqrt{71+12\sqrt{35}},\sqrt{71-12\sqrt{35}},-\sqrt{71-12\sqrt{35}}$

Step-by-step explanation:

Given : Equation $x^4-142x^2+1=0$

To find : The value of x?

Solution :

Let $y=x^2$ in the given equation $x^4-142x^2+1=0$

So, $(x^2)^2-142(x^2)+1=0$

$y^2-142y+1=0$

Solve by quadratic formula, $y=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Here, a=1, b=-142 and c=1

$y=\frac{-(-142)\pm\sqrt{(-142)^2-4(1)(1)}}{2(1)}$

$y=\frac{142\pm\sqrt{20164-4}}{2}$

$y=\frac{142\pm\sqrt{20160}}{2}$

$y=\frac{142\pm24\sqrt{35}}{2}$

$y=71\pm12\sqrt{35}$

i.e. $y=71+12\sqrt{35},71-12\sqrt{35}$

Now substitute in $y=x^2$

When $y=71+12\sqrt{35}$

$x^2=71+12\sqrt{35}$

$x=\pm\sqrt{71+12\sqrt{35}}$

When $y=71-12\sqrt{35}$

$x^2=71-12\sqrt{35}$

$x=\pm\sqrt{71-12\sqrt{35}}$

Therefore, the values of x are $x=\sqrt{71+12\sqrt{35}},-\sqrt{71+12\sqrt{35}},\sqrt{71-12\sqrt{35}},-\sqrt{71-12\sqrt{35}}$

Answer 2

Answer:

Step-by-step explanation: