(B) Let $|x-2| = y$. Since $|x-2| \ge 0$, we must have $y \ge 0$.Substituting into the equation, we get $y^2 + y - 2 = 0$.Factoring the quadratic equa

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(B) Let $|x-2| = y$. Since $|x-2| \ge 0$, we must have $y \ge 0$.Substituting into the equation, we get $y^2 + y - 2 = 0$.Factoring the quadratic equation: $(y+2)(y-1) = 0$.This gives $y = -2$ or $y = 1$.Since $y \ge 0$, we discard $y = -2$. Thus, $y = 1$.Now, solve $|x-2| = 1$:$x-2 = 1$ or $x-2 = -1$.$x = 3$ or $x = 1$.The real roots are $3$ and $1$.The sum of the roots is $3 + 1 = 4$.

Class 11 Mathematics 2 Complex Numbers and Quadratic Equations: Quadratic Equations Solution of quadratic equations and Nature of roots

Medium

The sum of all real roots of the equation $|x-2|^2+|x-2|-2=0$ is:

WBJEE 2009 All WBJEE

A
$7$
B
$4$
C
$1$
D
$5$

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2 Complex Numbers and Quadratic Equations: Quadratic Equations

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Solution of quadratic equations and Nature of roots

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