solve x^4-8x^3+18x^2-8x-7
Solve properly , I need an ans. worth 50 pts.
Answers
Answer:
x = 1 - √2
x = 3 - √2
x = 1 + √2
x = 3 + √2
Factorized answer = ( x² - 6 x + 7 )( x² - 2 x - 1 )
Step-by-step explanation:
For all the roots ( school method best )
x⁴ - 8 x³ + 18 x² - 8 x - 7 = 0
⇒ x⁴ - 2 x³ - x² - 6 x³ + 12 x² + 6 x + 7 x² - 14 x - 7 = 0
⇒ x² ( x² - 2 x - 1 ) - 6 x ( x² - 2 x - 1 ) + 7( x² - 2 x - 1 ) = 0
⇒ ( x² - 6 x + 7 )( x² - 2 x - 1 ) = 0
The factorized answer is required but I am also finding the roots of the given equation !
Apply Sridharacharya formula .
You will find :
x = 1 - √2
x = 3 - √2
x = 1 + √2
x = 3 + √2
Ignore this method . I was just experimenting with this ↓↓
Trial and error method
( quick for competitive exams [ but disadvantage is that it is applicable for finding one root only ] )
This is required when we are required to find only one root . Factorization takes a lot of time in such a case where the roots are rational .
There is no rational roots of the equation .We have to find the roots using the Bisection method .
Let f ( x ) = x⁴ - 8 x³ + 18 x² - 8 x - 7
Find the value of x such that f ( x ) is very close to zero .
x^4 - 8 x^3 + 18 x^2 - 8 x - 7
When x = 1
( 1 )^4 - 8 . 1 ^3 + 18 . 1^2 - 8.1 - 7
= > 1 - 8 + 36 - 8 - 7
= > 37 - 23
= > 14
When x = 0
( 0 )^4 - 8 . 0 ^3 + 18
0^2 - 8.0 - 7
= > - 7
When x = - 1
f ( x ) = 28
Hence x must be between - 1 and 0 .
When x = - 0.5
( -0.5 )^4 - 8 . ( -0.5 ) ^3 + 18
( -0.5 )^2 - 8 ( - 0.5 ) - 7
= > 2.5626 .
This means x > - 0.5
When x = - 0.414
f ( x ) = ( -0.414 )^4 - 8 . ( -0.414 ) ^3 + 18 . ( -0.414 )^2 - 8 ( - 0.414 ) - 7
= > -0.005
This very close to 0.Hence we consider this as an approximate solution .One solution is - 0.414. Very carefully we note that x = √2 = 1.414 and hence x - 1 = 0.414.
Now this can be written as - ( √2 - 1 ) ⇒ 1 - √2 .
The answer is hence 1 - √2 ( this is for any one root ).
Other roots may exist . But different approach is needed ..... sigh .
Step-by-step explanation:
plzz refer to the attached file..
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