Math, asked by tanzeelarehman52, 8 months ago

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

Answers

Answered by Anonymous
9

Answer:

142× x ×5 =20×5×142

x=20

Answered by ItzDeadDeal
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>

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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}}

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