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Answer 1

Question :

Solve the equation x⁴ - 10x³ + 26x² - 10x + 1 = 0

Answer :

Given equation :

x⁴ - 10x³ + 26x² - 10x + 1 = 0

To solve the equation let's convert the given equation into a quadratic equation.

Dividing throughout the equation by ' x² '

⇒ x⁴/x² - 10x³/x² + 26x²/x² - 10x/x² + 1/x² = 0

⇒ x² - 10x + 26 - 10/x + 1/x² = 0

Rearranging the terms so that we can take common

⇒ x² + 1/x² - 10x - 10/x  + 26 = 0

Taking ( - 10 ) common in the expression ( - 10x - 10/x ) we get,

⇒ x² + 1/x² - 10( x + 1/x ) + 26 = 0 → Eq( 1 )

To make the above equation into a quadratic equation we have to express ( x² + 1/x² ) in terms of ( x + 1/x )

We know that

a² + b² = ( a + b )² - 2ab

In the same way :

⇒ x² + 1/x² = ( x + 1/x )² - 2( x )( 1/x )

⇒ x² + 1/x² = ( x + 1/x )² - 2

Hence we have expressed ( x² + 1/x² ) in terms of ( x + 1/x ). So the Eq( 1 ) becomes

⇒ [ ( x + 1/x )² - 2 ]  - 10( x + 1/x ) + 26 = 0

Substituting ( x + 1/x ) = y in the above equation we get,

⇒  y² - 2 - 10y + 26 = 0

⇒ y² - 10y + 24 = 0

⇒ y² - 6y - 4y + 24 = 0

⇒ y( y - 6 ) - 4( y - 6 ) = 0

⇒ ( y - 6 )( y - 4 ) = 0

⇒ y - 6 = 0     OR     y - 4 = 0

Now we have two cases. Let's solve them one by one.

Case 1 :

⇒ y - 6 = 0

But x + 1/x = y

⇒ x + 1/x - 6 = 0

Multiply every term by ' x ' we get,

⇒ x² - 6x + 1 = 0

Using Quadratic formula

$\boxed{ \rm x =\dfrac{-b \pm\sqrt{b^2-4ac} }{2a} }$

Here we have,

• a = 1
• b = - 6
• c = 1

$\Rightarrow \sf x = \dfrac{-(-6)\pm\sqrt{(-6)^2-4(1)(1)} }{2(1)} \\\\\\ \Rightarrow \sf x = \dfrac{6\pm\sqrt{36-4} }{2} \\\\\\ \Rightarrow \sf x = \dfrac{6\pm\sqrt{32} }{2} \\\\\\ \Rightarrow \sf x = \dfrac{6 \pm 4\sqrt{2} }{2} \\\\\\ \Rightarrow \sf x = \dfrac{2( 3 \pm 2\sqrt{2} )}{2}$

⇒ x = 3 ± 2√2

Case 2 :

⇒ y - 4 = 0

But y = x + 1/x

⇒ x + 1/x - 4 =0

Multiplying every term with ' x ' we get,

⇒ x² - 4x + 1 = 0

Using Quadratic formula

$\boxed{ \rm x =\dfrac{-b \pm\sqrt{b^2-4ac} }{2a} }$

Here we have

• a = 1
• b = - 4
• c = 1

$\Rightarrow \sf x = \dfrac{-(-6)\pm\sqrt{(-4)^2-4(1)(1)} }{2(1)} \\\\\\ \Rightarrow \sf x = \dfrac{4 \pm\sqrt{16-4} }{2} \\\\\\ \Rightarrow \sf x = \dfrac{4 \pm \sqrt{12} }{2} \\\\\\ \Rightarrow \sf x = \dfrac{4 \pm 2\sqrt{3} }{2} \\\\\\ \Rightarrow \sf x = \dfrac{2( 2 \pm \sqrt{3} )}{2}$

⇒ x = 2 ± √3

∴ the roots of the given equation are ( 2 ± √3 ) and ( 3 ± 2√2 ).