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Quadratic Equation and Inequalities question

2003 · Shift 0 · Q102

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Quadratic Equation and Inequalities
2003 · Shift 0 · Q102

Quadratic Equation and Inequalities question

2003 · Shift 0 · Q102

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1

The number of real solutions of the equation

{x^2} - 3\left| x \right| + 2 = 0

is

A
3
B
2
C
4
D
1

View written solutionFree

Correct answer: C

We need to solve $x^2-3|x|+2=0.$
Since the equation involves $|x|$ , let $y=|x|.$ Then $y\ge 0$ and also $x^2=(|x|)^2=y^2.$ So the equation becomes $y^2-3y+2=0.$
Factor the quadratic: $y^2-3y+2=(y-1)(y-2)=0.$ Hence, $y=1 \quad \text{or} \quad y=2.$
Now substitute back $y=|x|$ :

If $|x|=1$ , then $x=\pm 1$ .
If $|x|=2$ , then $x=\pm 2$ .

Therefore the real solutions are $x=-2,-1,1,2,$ which gives a total of $4$ real solutions.
Checking options:

A: $3$ ❌
B: $2$ ❌
C: $4$ ✅
D: $1$ ❌

So the correct option is C.

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