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4-2.Quadratic Equations and Inequations
easy
If $|x - 2| + |x - 3| = 7$, then $x =$
A
$6$
B
$-1$
C
$6$ or $-1$
D
None of these
Solution
(c) Here $x = 2$ and $3$ are the critical points.
When $x < 2,|x – 2| = – (x – 2),|x – 3| = – (x – 3)$
$\therefore $ The given equation reduces to $2 – x + 3 – x = 7$
==> $x = – 1 < 2$
$\therefore $ $x = – 1$ is a solution.
When $2 \le x < 3,\,\,|x – 2| = x – 2,|x – 3| = – (x – 3)$
$\therefore $ The equation reduces to $x – 2 + 3 – x = 7$==> $1=7$
$\therefore $ No solution in this case.
When $x \ge 3$, the equation reduces to
$x – 2 + x – 3 = 7$ ==> $x = 6 > 3$
Hence we get, $x = 6$or $-1$
Trick : By inspection, we have that both the values $x = 6, – 1$ satisfy the given equation.
Standard 11
Mathematics
