Question

1
If x=
is the solution of the equation
1/5-2√6
(x^5 - 13x^4 + 38x^3 - 67x^2-53x +8 + k = 0), the value
of k is​

Answer 1

Answer:

plz make me a brain list

Step-by-step explanation:

Correct option is

A

96

Now

x

1

=5+2

6

Thus the value of (x−

x

1

)

2

=((5 - 2

6

) - (5 + 2

6

))

2

=(4\sqrt{6})^2= 16×6 = 96

Thus the correct answer is 96.

Answer 2

The value of k is -2.

Given:

x= $\frac{1}{5-2\sqrt{6} }$

$x^5 - 13x^4 + 38x^3 - 67x^2-53x +8 + k = 0$              

To find:

The value of k.

Solution:

x= $\frac{1}{5-2\sqrt{6} }$

$x^5 - 13x^4 + 38x^3 - 67x^2-53x +8 + k = 0$               ....................(1)

Rationalise  $\frac{1}{5-2\sqrt{6} }$.

x=  $\frac{1}{5-2\sqrt{6} }$× $\frac{5+2\sqrt{6} }{5+2\sqrt{6} }$

x=  $\frac{5+2\sqrt{6} }{(5)^2-(2\sqrt{6})^2 }$

x=  $\frac{5+2\sqrt{6} }{25-24 }$

x= 5+2√6

x-5= 2√6

Now, on squaring both the sides we get,

(x-5)²= (2√6)²

x²-10x+25=24

x²-10x+25-24=0

x²-10x+1=0                     ............................(2)

On dividing x²-10x+1 by $x^5 - 13x^4 + 38x^3 - 67x^2-53x +8 + k$ using long division method we get,

Quotient= x³-3x²+7x+6

Remainder= 2+k

We know that,

Dividend= Divisor×Quotient+Remainder

$x^5 - 13x^4 + 38x^3 - 67x^2-53x +8 + k$ = (x²-10x+1) × (x³-3x²+7x+6) + 2+k

From equation 1 and 2,

0= 0 × (x³-3x²+7x+6) + 2+k

0=2+k

k= -2

Therefore, the value of k is -2.

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