{x in R : |x - 2| = x^2} = | vedclass

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(D) The given equation is $|x - 2| = x^2$.Case $1$: If $x - 2 \ge 0$ (i.e.,$x \ge 2$),then $x - 2 = x^2$,which implies $x^2 - x + 2 = 0$. The discrimi

Vedclass · Accepted Answer

$\{1, -2\}$

Vedclass · Answer

(D) The given equation is $|x - 2| = x^2$.Case $1$: If $x - 2 \ge 0$ (i.e.,$x \ge 2$),then $x - 2 = x^2$,which implies $x^2 - x + 2 = 0$. The discriminant $D = (-1)^2 - 4(1)(2) = 1 - 8 = -7 Case $2$: If $x - 2 Factoring the quadratic equation: $x^2 + 2x - x - 2 = 0 \Rightarrow x(x + 2) - 1(x + 2) = 0 \Rightarrow (x - 1)(x + 2) = 0$.This gives $x = 1$ or $x = -2$.Since both $1 Therefore,the set of solutions is $\{1, -2\}$.

${x \in R : |x - 2| = x^2} = $

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