The solutions of the quadratic equation (3|x| | vedclass

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

Question

(D) The domain of definition of the function $y = \sqrt{x(x - 3)}$ is $x(x - 3) \ge 0$,which implies $x \le 0$ or $x \ge 3$. $(i)$The given equation i

Vedclass · Accepted Answer

$ -1/9, -2 $

Vedclass · Answer

(D) The domain of definition of the function $y = \sqrt{x(x - 3)}$ is $x(x - 3) \ge 0$,which implies $x \le 0$ or $x \ge 3$. $(i)$The given equation is $(3|x| - 3)^2 = |x| + 7$.Expanding the left side: $9|x|^2 - 18|x| + 9 = |x| + 7$.Rearranging the terms: $9|x|^2 - 19|x| + 2 = 0$.Let $t = |x|$,then $9t^2 - 19t + 2 = 0$.Factoring the quadratic: $(9t - 1)(t - 2) = 0$.Thus,$|x| = 1/9$ or $|x| = 2$.This gives $x = \pm 1/9$ or $x = \pm 2$.Checking these values against the domain condition $(i)$ ($x \le 0$ or $x \ge 3$):For $x = 2$,$2$ is not in the domain.For $x = -2$,$-2 \le 0$ (Valid).For $x = 1/9$,$1/9$ is not in the domain.For $x = -1/9$,$-1/9 \le 0$ (Valid).Therefore,the solutions are $-2$ and $-1/9$.

The solutions of the quadratic equation $(3|x| - 3)^2 = |x| + 7$ which belong to the domain of definition of the function $y = \sqrt{x(x - 3)}$ are given by

Similar Questions

The sum of the roots of the equation $x^2 + |2x - 3| - 4 = 0$ is:

If the roots of the equations $ax^2 + 2bx + c = 0$ and $bx^2 - 2\sqrt{ac}x + b = 0$ are real,then:

Let $a, b, c \in \mathbb{R}$ and $\alpha, \beta$ be the real roots of the equation $ax^2 + bx + c = 0$. If $a < 0, b > 0, c > 0$ and $\alpha < \beta$,then:

If the roots of the equation $(a^2 + b^2)x^2 - 2b(a + c)x + (b^2 + c^2) = 0$ are equal,then in which progression are $a, b, c$?

The roots of the equation $x^3-3x-2=0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module