The number of solutions of the equation left( | vedclass

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

Question

(B) Let $\frac{1}{\sqrt{x}} = \alpha$,where $x > 0$. Substituting this into the equation,we get: $(9\alpha^2 - 9\alpha + 2)(2\alpha^2 - 7\alpha + 3) =

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Let $\frac{1}{\sqrt{x}} = \alpha$,where $x > 0$. Substituting this into the equation,we get: $(9\alpha^2 - 9\alpha + 2)(2\alpha^2 - 7\alpha + 3) = 0$. Factoring the quadratic expressions: $(3\alpha - 1)(3\alpha - 2)(2\alpha - 1)(\alpha - 3) = 0$. This gives the roots for $\alpha$ as $\alpha = \frac{1}{3}, \frac{2}{3}, \frac{1}{2}, 3$. Since $\alpha = \frac{1}{\sqrt{x}}$,we have $\sqrt{x} = \frac{1}{\alpha}$,so $x = \frac{1}{\alpha^2}$. Calculating $x$ for each value of $\alpha$: For $\alpha = \frac{1}{3}, x = 9$. For $\alpha = \frac{2}{3}, x = \frac{9}{4}$. For $\alpha = \frac{1}{2}, x = 4$. For $\alpha = 3, x = \frac{1}{9}$. All these values of $x$ are positive and satisfy the condition $x > 0$. Thus,there are $4$ solutions.

The number of solutions of the equation $\left(\frac{9}{x}-\frac{9}{\sqrt{x}}+2\right)\left(\frac{2}{x}-\frac{7}{\sqrt{x}}+3\right)=0$ is :

Similar Questions

If $a, b, c, d$ are positive real numbers such that $a + b + c + d = 2$,then $M = (a + b)(c + d)$ satisfies the relation:

Let $f(x) = (1 + b^2)x^2 + 2bx + 1$ and $m(b)$ be the minimum value of $f(x)$ for a given $b$. As $b$ varies,the range of $m(b)$ is

If $a > 0$,then the value of $\sqrt{a + \sqrt{a + \sqrt{a + \dots \infty}}}$ is:

If $72^x \cdot 48^y = 6^{xy}$,where $x$ and $y$ are non-zero rational numbers,then $x+y$ equals

If $2, 1, 1$ are roots of the equation $x^3-4x^2+5x-2=0$,then find the roots of the equation $\left(x+\frac{1}{3}\right)^3-4\left(x+\frac{1}{3}\right)^2+5\left(x+\frac{1}{3}\right)-2=0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module