Question

question:-
Number of real solution(s) of x^2 - 4x+ 6 = 2 + root ( x-2) are-
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Answer 1

Question:-

• Number of real solution(s) of x² - 4x + 6 = 2 + root ( x-2) is ?

Solution:-

$\tt\implies \: { x }^{ 2 } - 4x + 4 + 2 = 2 + \sqrt { x - 2 }$

$\tt\implies \: { ( x - 2 ) }^{ 2 } =  \sqrt { x - 2 }$

Let ,

$\tt \: \sqrt { x - 2 } = x$

$\tt\implies \: { x }^{ 4 } - x = 0$

$\tt\implies \: x( { x }^{ 3 } - 1 ) = 0$

$\tt\implies \: x( x - 1 )( { x }^{ 2 } + x + 1 ) = 0$

$\tt\implies \: x = 0 , 1$

Answer 2

Answer:

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