If the set of all solutions of |x^2 + x - 9| | vedclass

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

Question

(B) We know that $|A + B| = |A| + |B|$ if and only if $AB \geq 0$.Given the equation $|x^2 + x - 9| = |x| + |x^2 - 9|$.Let $A = x$ and $B = x^2 - 9$.

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(B) We know that $|A + B| = |A| + |B|$ if and only if $AB \geq 0$.Given the equation $|x^2 + x - 9| = |x| + |x^2 - 9|$.Let $A = x$ and $B = x^2 - 9$. Then $A + B = x^2 + x - 9$.The condition for the equality to hold is $x(x^2 - 9) \geq 0$.Factoring the expression, we get $x(x - 3)(x + 3) \geq 0$.Using the wavy curve method (sign scheme), the solution set is $x \in [-3, 0] \cup [3, \infty)$.Comparing this with the given form $[\alpha, \beta] \cup [\gamma, \infty)$, we identify $\alpha = -3$, $\beta = 0$, and $\gamma = 3$.Therefore, the value of $(\alpha^2 + \beta^2 + \gamma^2) = (-3)^2 + 0^2 + 3^2 = 9 + 0 + 9 = 18$.

If the set of all solutions of $|x^2 + x - 9| = |x| + |x^2 - 9|$ is $[\alpha, \beta] \cup [\gamma, \infty)$, then $(\alpha^2 + \beta^2 + \gamma^2)$ is equal to:

Similar Questions

The region represented by the inequation $x-y \leq -1, x-y \geq 0, x \geq 0, y \geq 0$ is $.....$

The number of points $P(x, y)$ with natural numbers as coordinates that lie inside the quadrilateral formed by the lines $2x + y = 2$,$x = 0$,$y = 0$,and $x + y = 5$ is

The set of all real values of $x$ for which $\frac{x^2-1}{(x-4)(x-3)} \geq 1$ is

When $R$ is the set of all real numbers,$\{x \in R: \frac{\sqrt{12-x-x^2}}{x+10} \leq \frac{\sqrt{12-x-x^2}}{2x+9}\} = $

The set of values of $x$ which satisfy $5x + 2 < 3x + 8$ and $\frac{x + 2}{x - 1} < 4$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module