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A polynomial f(x) = x^4 -3x^3 -2x^2 -3x +1 = 0, has at most 4 roots. Find the sum of all the real roots.
Hint: Try to convert it into a quadratic equation 1st.
Answers
Answer:
4
Explanation:
x^4 - 3x³ - 2x² - 3x + 1 = 0
Dividing throughout by ' x² '
⇒ ( x^4 - 3x³ - 2x² - 3x + 1 ) / x² = 0
⇒ x^4/x² - 3x³/x² - 2x²/x² - 3x/x² + 1/x² = 0
⇒ x² - 3x - 2 - 3/x + 1/x² = 0
Rearranging the terms
⇒ x² + 1/x² - 3x - 3/x - 2 = 0
⇒ ( x² + 1/x² ) - 3( x + 1/x ) - 2 = 0
x² + 1/x² can be written as { ( x + 1/x )² - 2 }
⇒ { ( x + 1/x )² - 2 } - 3( x + 1/x ) - 2 = 0
Substituting ( x + 1/x ) = y in the above equation
⇒ y² - 2 - 3y - 2 = 0
⇒ y² - 3y - 4 = 0
⇒ y² - 4y + y - 4 = 0
⇒ y( y - 4 ) + 1( y - 4 ) = 0
⇒ ( y - 4 )( y + 1 ) = 0
⇒ y - 4 = 0 or y + 1 = 0
Case 1 :
⇒ y - 4 = 0
⇒ x + 1/x - 4 = 0
⇒ x² - 4x + 1 = 0
⇒ x = ( 4 ± 2√3 ) / 2
⇒ x = { 2( 2 ± √3 ) } / 2
⇒ x = 2 ± √3
⇒ x = 2 + √3 or x = 2 - √3
Case 2 :
⇒ y + 1 = 0
⇒ x + 1/x + 1 = 0
⇒ x² + x + 1 = 0
⇒ x = ( - 1 ± √3 i ) / 2
Therefore the given equation has 2 real roots i.e 2 + √3 and 2 - √3.
Sum of all real roots = 2 + √3 + 2 - √3 = 4
Hence the sum of all real roots is 4.
Answer:
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