Question

3. Find the value of x in the following:
a 142 x x x 5 = 20 x 5 x 142​

Answer 1

Answer:

142× x ×5 =20×5×142

x=20

Answer 2

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=0x

4

−142x

2

+1=0

To find : The value of x?

Solution :

Let y=x^2y=x

2

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

4

−142x

2

+1=0

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

2

)

2

−142(x

2

)+1=0

y^2-142y+1=0y

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

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

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

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

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

$y=71\pm12\sqrt{35}y</p><p>$

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

Now substitute in y=x^2y=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

$x=\sqrt{71+12\sqrt{35}},-\sqrt{71+12\sqrt{35}},\sqrt{71-12\sqrt{35}},-\sqrt{71-12\sqrt{35}}$