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Answers
Step-by-step explanation:
Given :-
x = 1/(5-2√6) is the solution of the equation x⁵-13x⁴+38x³-67x²-53x+8+k = 0
To find :-
Find the value of k ?
Solution :-
Given equation is
x⁵-13x⁴+38x³-67x²-53x+8+k = 0
Let p(x) = x⁵-13x⁴+38x³-67x²-53x+8+k = 0
Given solution of p(x) = 1/(5-2√6)
The Rationalising factor of 5-2√6 is 5+2√6
On Rationalising the denominator then
=> x = [1/(5-2√6)]×[(5+2√6)/(5+2√6)]
=> x = (5+2√6)/[(5)²-(2√6)²]
=> x = (5+2√6)/(25-24)
=> x = 5+2√6
=> x-5 = 2√6
On squaring both sides then
=> (x-5)² = (2√6)²
=> x²-10x+25 = 24
=> x²-10x+25-24 = 0
=>x²-10x+1 = 0
So now , on dividing p(x) by x²-10x+1 then
x³-3x²+7x+6
_____________________
x²-10x+1)x⁵-13x⁴+38x³-67x²-53x+8+k
x⁵-10x⁴+x³
(-) (+) (-)
______________________
-3x⁴+37x³-67x²
-3x⁴+30x³-3x²
(+) (-) (+)
_______________________
7x³ -64x²-53x
7x³-70x²+7x
(-) (+) (-)
________________________
6x²-60x+8+k
6x²-60x+6
(-) (+) (-)
_________________________
2+k
_________________________
We get Quotient = x³-3x²+7x+6
Remainder = 2+k
But x is the solution of P(x).
By Factor theorem the remainder must be equal to zero
=> 2+k = 0
=> 2 = -k
=> -2 = k
Therefore, k = -2
Answer:-
The value of k for the given problem is -2
Used formulae:-
- Rationalising factor of a-√b is a+√b
- Let P(x) be a polynomial of the degree greater than or equal to 1 and x-a is another linear polynomial, if x-a is a factor of P (x) then P(a) = 0 is called Factor Theorem.
Answer:
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