If |x - 2| + |x - 3| = 7,then x = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The critical points for the given equation are $x = 2$ and $x = 3$.
Case $1$: If $x < 2$,then $|x - 2| = -(x - 2)$ and $|x - 3| = -(x - 3)$.

Vedclass · Accepted Answer

$6$ or $-1$

Vedclass · Answer

(C) The critical points for the given equation are $x = 2$ and $x = 3$.
Case $1$: If $x < 2$,then $|x - 2| = -(x - 2)$ and $|x - 3| = -(x - 3)$.
The equation becomes $-(x - 2) - (x - 3) = 7$,which simplifies to $-x + 2 - x + 3 = 7$,or $-2x + 5 = 7$.
$-2x = 2$,so $x = -1$. Since $-1 < 2$,this is a valid solution.
Case $2$: If $2 \le x < 3$,then $|x - 2| = x - 2$ and $|x - 3| = -(x - 3)$.
The equation becomes $(x - 2) - (x - 3) = 7$,which simplifies to $x - 2 - x + 3 = 7$,or $1 = 7$.
This is a contradiction,so there is no solution in this interval.
Case $3$: If $x \ge 3$,then $|x - 2| = x - 2$ and $|x - 3| = x - 3$.
The equation becomes $(x - 2) + (x - 3) = 7$,which simplifies to $2x - 5 = 7$,or $2x = 12$.
Thus,$x = 6$. Since $6 \ge 3$,this is a valid solution.
Therefore,the solutions are $x = 6$ or $x = -1$.

If $|x - 2| + |x - 3| = 7$,then $x =$

Similar Questions

The roots of the equation $x^2 + bx - c = 0$ where $b, c > 0$ are:

If $\alpha$ and $\beta$ are the roots of the equation ${x^2} - a(x + 1) - b = 0$,then $(\alpha + 1)(\beta + 1) = $

If $x+\frac{1}{y}=1$ and $y+\frac{1}{z}=1$,then find the value of $z+\frac{1}{x}$.

If $\alpha$ and $\beta$ are the roots of the equation $2x^2 - 35x + 2 = 0$,then the value of $(2\alpha - 35)^3 \cdot (2\beta - 35)^3$ is equal to

Solve the given two equations and select the correct answer from the given options.
$I. x^{2}-2x-15=0$
$II. y^{2}+5y+6=0$

Vedclass Test Series

Exam Paper Generator

Online Exam Module