(D) Given the equation $|x - 2| = x^2$. Case $1$: If $x - 2 \ge 0$ (i.e.,$x \ge 2$),then $x - 2 = x^2 \Rightarrow x^2 - x + 2 = 0$. The discriminant $

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(D) Given the equation $|x - 2| = x^2$. Case $1$: If $x - 2 \ge 0$ (i.e.,$x \ge 2$),then $x - 2 = x^2 \Rightarrow x^2 - x + 2 = 0$. The discriminant $D = (-1)^2 - 4(1)(2) = 1 - 8 = -7 Case $2$: If $x - 2 Factoring the quadratic equation: $(x + 2)(x - 1) = 0$. This gives $x = -2$ or $x = 1$. Both values satisfy the condition $x Thus,the solution set is $\{ 1, -2 \}$.

Class 11 Mathematics 4-2.Quadratic Equations and Inequations Solution of quadratic equations and Nature of roots

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Find the set $\{ x \in R : |x - 2| = x^2 \}$.

A
$\{ -1, 2 \}$
B
$\{ 1, 2 \}$
C
$\{ -1, -2 \}$
D
$\{ 1, -2 \}$

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4-2.Quadratic Equations and Inequations

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Solution of quadratic equations and Nature of roots

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Find the set $\{ x \in R : |x - 2| = x^2 \}$.

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If the roots of the equation $\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}$ are equal in magnitude but opposite in sign,then $r = ......$

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