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 one of the roots of the equation $6x^3-25x^2+2x+8=0$ is an integer and $\alpha > 0$,$\beta < 0$ are the other two roots,then $\frac{4}{\alpha}+\frac{1}{\beta}=$

If the two roots of the equation $(a - 1)(x^4 + x^2 + 1) + (a + 1)(x^2 + x + 1)^2 = 0$ are real and distinct,then the set of all values of $a$ is

The sum of all the real roots of the equation $|x-2|^2+|x-2|-2=0$ is

The roots of the quadratic equation $2x^2 + 3x + 1 = 0$ are:

The number of distinct real roots of the equation $x^{5}(x^{3}-x^{2}-x+1)+x(3x^{3}-4x^{2}-2x+4)-1=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module