For the equation |x^2| + |x| - 6 = 0,the root | vedclass

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

Question

(C) Given equation: $|x^2| + |x| - 6 = 0$.Since $|x^2| = |x|^2$,the equation can be written as $|x|^2 + |x| - 6 = 0$.Let $t = |x|$,where $t \ge 0$. Th

Vedclass · Accepted Answer

Real with sum zero

Vedclass · Answer

(C) Given equation: $|x^2| + |x| - 6 = 0$.Since $|x^2| = |x|^2$,the equation can be written as $|x|^2 + |x| - 6 = 0$.Let $t = |x|$,where $t \ge 0$. The equation becomes $t^2 + t - 6 = 0$.Factoring the quadratic: $(t + 3)(t - 2) = 0$.This gives $t = -3$ or $t = 2$.Since $t = |x| \ge 0$,we discard $t = -3$.Thus,$|x| = 2$,which implies $x = 2$ or $x = -2$.The roots are $2$ and $-2$.The sum of the roots is $2 + (-2) = 0$.

For the equation $|x^2| + |x| - 6 = 0$,the roots are

Similar Questions

Let $P(x) = x^3 - ax^2 + bx + c$ where $a, b, c \in \mathbb{R}$ have integral roots such that $P(6) = 3$. Then $a$ cannot be equal to:

The value of $4+\frac{1}{5+\frac{1}{4+\frac{1}{5+\frac{1}{4+\ldots \ldots \infty}}}}$ is

The maximum possible number of real roots of the equation $x^5 - 6x^2 - 4x + 5 = 0$ is

If the roots of $x^2 - bx + c = 0$ are two consecutive integers,then $b^2 - 4c$ is

The sum of all non-integer roots of the equation $x^5-6x^4+11x^3-5x^2-3x+2=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module