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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
[tex]\boxed{ \rm x =\dfrac{-b \pm\sqrt{b^2-4ac} }{2a} }[/tex]
Here we have,
[tex]\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}[/tex]
⇒ 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
[tex]\boxed{ \rm x =\dfrac{-b \pm\sqrt{b^2-4ac} }{2a} }[/tex]
Here we have
[tex]\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}[/tex]
⇒ x = 2 ± √3
∴ the roots of the given equation are ( 2 ± √3 ) and ( 3 ± 2√2 ).