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Answers
Answer:
x = -1, 3, 1 ± √2
Step-by-step explanation:
Given Equation is (x² - 2x)² - 4(x² - 2x) + 3 = 0
⇒ x⁴ + 4x² - 4x³ - 4x² + 8x + 3 = 0
⇒ x⁴ - 4x³ + 8x + 3 = 0
⇒ x⁴ - 5x³ + x³ + 5x² - 5x² + 3x + 5x + 3 = 0
⇒ x⁴ - 5x³ + 5x² + 3x + x³ - 5x² + 5x + 3 = 0
⇒ x(x³ - 5x² + 5x + 3) + 1(x³ - 5x² + 5x + 3) = 0
⇒ (x + 1)(x³ - 5x² + 5x + 3) = 0
⇒ (x + 1)[x³ - 2x² - 3x² + 6x - x + 3] = 0
⇒ (x + 1)[x³ - 2x² - x - 3x² + 6x + 3] = 0
⇒ (x + 1)[x(x² - 2x - 1) - 3(x² - 2x - 1)] = 0
⇒ (x + 1)[(x - 3)(x² - 2x - 1)] = 0
⇒ (x + 1)(x - 3)(x² - 2x - 1) = 0
(i)
x + 1 = 0
x = -1
(ii)
x - 3 = 0
x = 3.
(iii)
x² - 2x - 1 = 0
D = b² - 4ac
= (-2)² - 4(1)(-1)
= 4 + 4
= 8.
The solutions are:
x = -b ± √D/2a
= -(-2) ± √8/2
= 2 ± √8/2
= 1 ± √2/2
Therefore, the values are:
x = -1,3, 1±√2.
Hope it helps!
( x² - 2 x )² - 4 ( x² - 2 x ) + 3 = 0
Let x² - 2 x be k
==> k² - 4 k + 3 = 0
Now easy na??
==> k² - 3 k - k + 3 = 0
==> k ( k - 3 ) - 1 ( k - 3 ) = 0
==> ( k - 3 )( k - 1 ) = 0
Either :
k - 3 = 0 ==> k = 3
or :
k - 1 = 0 ==> k = 1
When k = 3
x² - 2 x = 3
==> x² - 2 x - 3 = 0
==> x² - 3 x + x - 3 = 0
==> x ( x - 3 ) + 1 ( x - 3 ) = 0
==> ( x + 1 )( x - 3 ) = 0
Either :
x + 1 = 0
==> x = - 1
or :
x - 3 = 0
==> x = 3
When k = 1
x² - 2 x = 1
==> x² - 2 x - 1 = 0
Comparing with a x² + b x + c = 0
a = 1
b = - 2
c = - 1
[ By quadratic formula ]
==> x = [ - b ± √ ( b² - 4 ac ) ] / 2 a
==> x = [ 2 ± √ ( 4 - 4 × - 1 ) ] / 2 × 1
==> x = [ 2 ± √8 ] / 2
==> x = [ 2 ± 2√2 ] /2
==> x = 2 [ 1 ± √2 ] / 2
==> x = 1 ± √2
ANSWER
x = - 1 ,
x = 3 ,
x = 1 + √2 ,
x = 1 - √2 .
Hope it helps buddy :-)
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