Question

Number of real solution(s) of x^2 - 4x+ 6 = 2 + root ( x-2) is /are ?

Answer 1

Sol.

Let $\bold{f(x)=x^2-4x+6}$ and $\bold{g(x)=2+\sqrt{x-2}}$.

If two graph intersects, that means $f(x)=g(x)$.

$f(x)=g(x)$

$\Longleftrightarrow (x^2-4x+4)+2=2+\sqrt{x-2}$

$\Longleftrightarrow (x-2)^2=\sqrt{x-2}$

Let $\sqrt{x-2}=t$ and solve the equation.

$t^4-t=0$

$\Longleftrightarrow t(t^3-1)=0$

$\Longleftrightarrow t(t-1)(t^2+t+1)=0$

$t=0,1$ is the solution of the equation. So, they intersect twice.

The number of real solutions is 2.

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Inverse Function

$y=(x-2)^2$ for $x\geq 2$ (Positive number set.)

Change both domain and range.

$x=(y-2)^2$

Solve for $y$ to simplify the inverse function.

$y-2=\sqrt{x}$

$\therefore y=\sqrt{x} +2$

Surprisingly, the intersection of the previous function and the inverse lie on $y=x$. (Increasing functions.)

Answer 2

$\huge\underline{\underline {\sf{Question}}} :$

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

$\huge\underline{\underline {\sf{Solution}}} :$

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

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

Let ,

• $\tt \: \sqrt { x - 2 } = x$

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

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

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

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