What is the number of real solutions for the | vedclass

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

Question

(A) Given the equation $x^2 - 3|x| + 2 = 0$.Since $x^2 = |x|^2$,we can rewrite the equation as $|x|^2 - 3|x| + 2 = 0$.Let $t = |x|$,where $t \geq 0$.

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) Given the equation $x^2 - 3|x| + 2 = 0$.Since $x^2 = |x|^2$,we can rewrite the equation as $|x|^2 - 3|x| + 2 = 0$.Let $t = |x|$,where $t \geq 0$. The equation becomes $t^2 - 3t + 2 = 0$.Factoring the quadratic equation: $(t - 1)(t - 2) = 0$.This gives $t = 1$ or $t = 2$.Case $1$: $|x| = 1$ implies $x = 1$ or $x = -1$.Case $2$: $|x| = 2$ implies $x = 2$ or $x = -2$.Thus,the real solutions are $x \in \{-2, -1, 1, 2\}$.The total number of real solutions is $4$.

What is the number of real solutions for the equation $x^2 - 3|x| + 2 = 0$?

Similar Questions

If $r$ and $s$ are positive,what is the nature of the roots of the quadratic equation $ax^2 - rx - s = 0$?

The value of $\sqrt{42+\sqrt{42+\sqrt{42+\ldots}}}$ is equal to:

Solve the equation $\frac{p + q - x}{r} + \frac{q + r - x}{p} + \frac{r + p - x}{q} + \frac{4x}{p + q + r} = 0$.

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3+3x^2-x-3=0$,then $(1+\alpha^2)(1+\beta^2)(1+\gamma^2) = $

If the roots of the equation $x^2 + px + q = 0$ are $\alpha$ and $\beta$,and the roots of the equation $x^2 - xr + s = 0$ are $\alpha^4$ and $\beta^4$,then the roots of the equation $x^2 - 4qx + 2q^2 - r = 0$ will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module